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Answer:

Yes, 0 is a rational number.

0 can be expressed in the form of the fraction $\frac{p}{q}$, where $p=0$ and q can be any integer except 0.

Answer:

(i) $\frac{5}{7}$

(ii) $\frac{8}{3}$
$\frac{8}{3}=2\frac{2}{3}$

(iii) $-\frac{23}{6}=-3\frac{5}{6}$

(iv) 1.3
$1.3=\frac{13}{10}=1\frac{3}{10}$

(v) – 2.4
$-2.4=\frac{-24}{10}=\frac{-12}{5}=-2\frac{2}{5}$

Answer:

(i)
Let:
x = $\frac{3}{8}$ and y = $\frac{2}{5}$
Rational number lying between x and y:

=

(ii) 1.3 and 1.4
Let:
x = 1.3 and y = 1.4
Rational number lying between x and y:

=

(iii)
Let:
x = $-$1 and y = $\frac{1}{2}$
Rational number lying between x and y:

= $-\frac{1}{4}$

(iv)
Let:
x$-\frac{3}{4}$ and y = $-\frac{2}{5}$
Rational number lying between x and y:

=

(v)
A rational number lying between  will be
$\frac{1}{2}\left(\frac{1}{9}+\frac{2}{9}\right)=\frac{1}{2}×\frac{1}{3}=\frac{1}{6}$

Answer:

n = 3
$d=\frac{\left(y-x\right)}{n+1}=\frac{\frac{7}{8}-\frac{3}{5}}{3+1}=\frac{11}{40}×\frac{1}{4}=\frac{11}{160}$
Rational numbers between  will be
$\left(x+d\right),\left(x+2d\right),...,\left(x+nd\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{3}{5}+\frac{11}{160}\right),\left(\frac{3}{5}+2×\frac{11}{160}\right),\left(\frac{3}{5}+3×\frac{11}{160}\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{107}{160}\right),\left(\frac{118}{160}\right),\left(\frac{129}{160}\right)\phantom{\rule{0ex}{0ex}}⇒\left(\frac{107}{160}\right),\left(\frac{59}{80}\right),\left(\frac{129}{160}\right)$
There are infinitely many rational numbers between two given rational numbers.

Answer:

n = 4
n + 1 = 4 + 1 = 5

Thus, rational numbers between  are $\frac{16}{35},\frac{17}{35},\frac{18}{35},\frac{19}{35}$.

Answer:

x = 2, y = 3 and n = 6
$d=\frac{y-x}{n+1}=\frac{3-2}{6+1}=\frac{1}{7}$
Thus, the required numbers are
$\left(x+d\right),\left(x+2d\right),\left(x+3d\right),...,\left(x+nd\right)\phantom{\rule{0ex}{0ex}}=\left(2+\frac{1}{7}\right),\left(2+2×\frac{1}{7}\right),\left(2+3×\frac{1}{7}\right),\left(2+4×\frac{1}{7}\right),\left(2+5×\frac{1}{7}\right),\left(2+6×\frac{1}{7}\right)\phantom{\rule{0ex}{0ex}}=\frac{15}{7},\frac{16}{7},\frac{17}{7},\frac{18}{7},\frac{19}{7},\frac{20}{7}$

Answer:

n = 5
n + 1 = 6
$x=\frac{3}{5},y=\frac{2}{3}$
$d=\frac{y-x}{n+1}=\frac{\frac{2}{3}-\frac{3}{5}}{6}=\frac{10-9}{90}=\frac{1}{90}$

Thus, rational numbers between  will be
$\left(x+d\right),\left(x+2d\right),\left(x+3d\right),\left(x+4d\right),\left(x+5d\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{3}{5}+\frac{1}{90}\right),\left(\frac{3}{5}+\frac{2}{90}\right),\left(\frac{3}{5}+\frac{3}{90}\right),\left(\frac{3}{5}+\frac{4}{90}\right),\left(\frac{3}{5}+\frac{5}{90}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{55}{90}\right),\left(\frac{56}{90}\right),\left(\frac{57}{90}\right),\left(\frac{58}{90}\right),\left(\frac{59}{90}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{11}{18}\right),\left(\frac{28}{45}\right),\left(\frac{19}{30}\right),\left(\frac{29}{45}\right),\left(\frac{59}{90}\right)\phantom{\rule{0ex}{0ex}}$

Answer:

Let:
x = 2.1, y = 2.2 and n = 16

We know:
d = $\frac{y-x}{n+1}=\frac{2.2-2.1}{16+1}=\frac{0.1}{17}=\frac{1}{170}$= 0.005 (approx.)
So, 16 rational numbers between 2.1 and 2.2 are:
(x + d), (x + 2d), ...(x + 16d)
= [2.1 + 0.005], [2.1 + 2(0.005)],...[2.1 + 16(0.005)]
= 2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17, 2.175 and 2.18

Answer:

(i) Every natural number is a whole number.
True, since natural numbers are counting numbers i.e N = 1, 2,...
Whole numbers are natural numbers together with 0. i.e W = 0, 1, 2,...
So, every natural number is a whole number

(ii) Every whole number is a natural number.
False, as whole numbers contain natural numbers and 0 whereas natural numbers only contain the counting numbers except 0.

(iii) Every integer is a whole number.
False, whole numbers are natural numbers together with a zero whereas integers include negative numbers also.

(iv) Every integer is a rational number.
True, as rational numbers are of the form . All integers can be represented in the form .
(v) Every rational number is an integer.
False, as rational numbers are of the form . Integers are negative and positive numbers which are not in $\frac{p}{q}$ form.
For example, $\frac{1}{2}$ is a rational number but not an integer.

(vi) Every rational number is a whole number.
False, as rational numbers are of the form . Whole numbers are natural numbers together with a zero.
For example, $\frac{5}{7}$ is a rational number but not a whole number.

Answer:

(i) $\frac{13}{80}$
Denominator of $\frac{13}{80}$ is 80.
And,
80 = 24$×$5
Therefore, 80 has no other factors than 2 and 5.
Thus, $\frac{13}{80}$ is a terminating decimal.

(ii) $\frac{7}{24}$
Denominator of $\frac{7}{24}$ is 24.
And,
24 = 23$×$3
So, 24 has a prime factor 3, which is other than 2 and 5.
Thus, $\frac{7}{24}$ is not a terminating decimal.

(iii) $\frac{5}{12}$
Denominator of $\frac{5}{12}$ is 12.
And,
12 = 22$×$3
So, 12 has a prime factor 3, which is other than 2 and 5.
Thus, $\frac{5}{12}$ is not a terminating decimal.

(iv) $\frac{31}{375}$
Denominator of $\frac{31}{375}$ is 375.
$375={5}^{3}×3$
So, the prime factors of 375 are 5 and 3.
Thus, $\frac{31}{375}$ is not a terminating decimal.

(v) $\frac{16}{125}$
Denominator of $\frac{16}{125}$ is 125.
And,
125 = 53
Therefore, 125 has no other factors than 2 and 5.
Thus, $\frac{16}{125}$ is a terminating decimal.

Answer:

(i) $\frac{5}{8}$ = 0.625
By actual division, we have:

It is a terminating decimal expansion.

(ii) $\frac{7}{25}$
$\frac{7}{25}$ = 0.28
By actual division, we have:

It is a terminating decimal expansion.

(iii) $\frac{3}{11}$ = $0.\overline{27}$

It is a non-terminating recurring decimal.
(iv) $\frac{5}{13}$ = $0.\overline{384615}$

It is a non-terminating recurring decimal.

(v) $\frac{11}{24}$
$\frac{11}{24}$ =
By actual division, we have:

It is nonterminating recurring decimal expansion.

(vi) $\frac{261}{400}$$=0.6525$

It is a terminating decimal expansion.

(vii) $\frac{231}{625}$$=0.3696$

It is a terminating decimal expansion.

(viii) $2\frac{5}{12}$
2$\frac{5}{12}$ = $\frac{29}{12}$ =
By actual division, we have:

It is non-terminating decimal expansion.

Answer:

(i) $0.\overline{2}$
Let x = 0.222...                               .....(i)
Only one digit is repeated so, we multiply x by 10.
10x = 2.222...                                 .....(ii)
Subtracting (i) from (ii) we get
$9x=2\phantom{\rule{0ex}{0ex}}⇒x=\frac{2}{9}$

(ii) $0.\overline{53}$
Let x = 0.5353...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 53.5353...                                 .....(ii)
Subtracting (i) from (ii) we get
$99x=53\phantom{\rule{0ex}{0ex}}⇒x=\frac{53}{99}$

(iii) $2.\overline{93}$
Let x = 2.9393...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 293.9393...                                 .....(ii)
Subtracting (i) from (ii) we get
$99x=291\phantom{\rule{0ex}{0ex}}⇒x=\frac{291}{99}=\frac{97}{33}$

(iv) $18.\overline{48}$
Let x = 18.4848...                               .....(i)
Two digits are repeated so, we multiply x by 100.
100x = 1848.4848...                                 .....(ii)
Subtracting (i) from (ii) we get
$99x=1830\phantom{\rule{0ex}{0ex}}⇒x=\frac{1830}{99}=\frac{610}{33}$

(v) $0.\overline{235}$
Let x = 0.235235...                               .....(i)
Three digits are repeated so, we multiply x by 1000.
1000x = 235.235235...                                 .....(ii)
Subtracting (i) from (ii) we get
$999x=235\phantom{\rule{0ex}{0ex}}⇒x=\frac{235}{999}$

(vi) $0.00\overline{32}$
Let x = 0.003232...                               .....(i)
we multiply x by 100.
100x = 0.3232...                                 .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 32.3232...                           .....(iii)
Subtracting (ii) from (iii) we get
$9900x=32\phantom{\rule{0ex}{0ex}}⇒x=\frac{32}{9900}=\frac{8}{2475}$

(vii) $1.3\overline{23}$
Let x = 1.32323...                               .....(i)
we multiply x by 10.
10x = 13.2323...                                 .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
1000x = 1323.2323...                           .....(iii)
Subtracting (ii) from (iii) we get
$990x=1310\phantom{\rule{0ex}{0ex}}⇒x=\frac{131}{99}$

(viii) $0.3\overline{178}$
Let x = 0.3178178...                               .....(i)
we multiply x by 10.
10x = 3.178178...                                 .....(ii)
Again multiplying by 1000 as there are 3 repeating numbers after decimals we get
10000x = 3178.178178...                           .....(iii)
Subtracting (ii) from (iii) we get
$9990x=3175\phantom{\rule{0ex}{0ex}}⇒x=\frac{3175}{9990}=\frac{635}{1998}$

(ix) $32.12\overline{35}$
Let x = 32.123535...                               .....(i)
we multiply x by 100.
100x = 3212.3535...                                 .....(ii)
Again multiplying by 100 as there are 2 repeating numbers after decimals we get
10000x = 321235.35...                           .....(iii)
Subtracting (ii) from (iii) we get
$9900x=318023\phantom{\rule{0ex}{0ex}}⇒x=\frac{318023}{9900}$

(x) $0.40\overline{7}$
Let x = 0.40777...                               .....(i)
we multiply x by 100.
100x = 40.7777...                                 .....(ii)
Again multiplying by 10 as there is 1 repeating number after decimals we get
1000x = 407.777...                           .....(iii)
Subtracting (ii) from (iii) we get
$900x=367\phantom{\rule{0ex}{0ex}}⇒x=\frac{367}{900}$

Answer:

Given: $2.\overline{36}+0.\overline{23}$
Let

First we take x and convert it into $\frac{p}{q}$
100x = 236.3636...        ...(iii)
Subtracting (i) from (iii) we get
$99x=234\phantom{\rule{0ex}{0ex}}⇒x=\frac{234}{99}$
Similarly, multiply y with 100 as there are 2 decimal places which are repeating themselves.
$100y=23.2323...$               ...(iv)
Subtracting (ii) from (iv) we get
$99y=23\phantom{\rule{0ex}{0ex}}⇒y=\frac{23}{99}$
Adding x and y we get
$2.\overline{36}+0.\overline{23}$$x+y=\frac{234}{99}+\frac{23}{99}=\frac{257}{99}$

Answer:

x = 0.3838...                                  ...(i)
Multiply with 100 as there are 2 repeating digits after decimals
100x = 38.3838...                          ...(ii)
Subtracting (i) from (ii) we get
99x = 38
$⇒x=\frac{38}{99}$
Similarly, we take
y = 1.2727...                                  ...(iii)
Multiply y with 100 as there are 2 repeating digits after decimal.
100y = 127.2727...                       ...(iv)
Subtract (iii) from (iv) we get
99y = 126

Answer:

A number that can neither be expressed as a terminating decimal nor be expressed as a repeating decimal is called an irrational number. A rational number, on the other hand, is always a terminating decimal, and if not, it is a repeating decimal.
Examples of irrational numbers:
0.101001000...
0.232332333...

Answer:

(i) $\sqrt{\frac{3}{81}}$
$\sqrt{\frac{3}{81}}=\sqrt{\frac{1}{27}}=\frac{1}{3}\sqrt{\frac{1}{3}}$
It is an irrational number.

(ii) $\sqrt{361}$ = 19
So, it is rational.

(iii) $\sqrt{21}$

(iv) $\sqrt{1.44}$ = 1.2
So, it is rational.

(v) $\frac{2}{3}\sqrt{6}$
It is an irrational number

(vi) 4.1276
It is a terminating decimal. Hence, it is rational.

(vii) $\frac{22}{7}$

(viii) 1.232332333...

(xi) 6.834834... is a rational number because it is repeating.

Answer:

x be a rational number and y be an irrational number then x + y necessarily will be an irrational number.
Example: 5 is a rational number but $\sqrt{2}$ is irrational.
So, 5 + $\sqrt{2}$ will be an irrational number.

Answer:

a be a rational number and b be an irrational number then ab necessarily will be an irrational number.
Example: 6 is a rational number but $\sqrt{5}$ is irrational. And 6$\sqrt{5}$ is also an irrational number.

Answer:

Product of two irrational numbers is not always an irrational number.
Example: $\sqrt{5}$ is irrational number. And $\sqrt{5}×\sqrt{5}=5$ is a rational number. But the product of another two irrational numbers is $\sqrt{6}$ which is also an irrational numbers.

Answer:

(i) 2 irrational numbers with difference an irrational number will be .
(ii) 2 irrational numbers with difference is a rational number will be
(iii) 2 irrational numbers with sum an irrational number
(iv) 2 irrational numbers with sum a rational number is
(v) 2 irrational numbers with product an irrational number will be
(vi) 2 irrational numbers with product a rational number will be
(vii) 2 irrational numbers with quotient an irrational number will be
(viii) 2 irrational numbers with quotient a rational number will be .

Answer:

(i) Let us assume, to the contrary, that $3+\sqrt{3}$ is rational.
Then, $3+\sqrt{3}=\frac{p}{q}$, where p and q are coprime and $q\ne 0$.
$⇒\sqrt{3}=\frac{p}{q}-3\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}=\frac{p-3q}{q}$
Since, p and q are are integers.
$⇒\frac{p-3q}{q}$ is rational.
So, $\sqrt{3}$ is also rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $3+\sqrt{3}$ is rational.
Hence, $3+\sqrt{3}$ is irrational.

(ii) Let us assume, to the contrary, that $\sqrt{7}-2$ is rational.
Then, $\sqrt{7}-2=\frac{p}{q}$, where p and q are coprime and $q\ne 0$.
$⇒\sqrt{7}=\frac{p}{q}+2\phantom{\rule{0ex}{0ex}}⇒\sqrt{7}=\frac{p+2q}{q}$
Since, p and q are are integers.
$⇒\frac{p+2q}{q}$ is rational.
So, $\sqrt{7}$ is also rational.
But this contradicts the fact that $\sqrt{7}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $\sqrt{7}-2$ is rational.
Hence, $\sqrt{7}-2$ is irrational.

(iii) As, $\sqrt[3]{5}×\sqrt[3]{25}$

Hence, $\sqrt[3]{5}×\sqrt[3]{25}$ is rational.

(iv) As, $\sqrt{7}×\sqrt{343}$

Hence, $\sqrt{7}×\sqrt{343}$ is rational.

(v) As,
Hence, $\sqrt{\frac{13}{117}}$ is rational.

(vi) As, $\sqrt{8}×\sqrt{2}$

Hence, $\sqrt{8}×\sqrt{2}$ is rational.

Answer:

As, few rational numbers between 2 and 2.5 are: 2.1, 2.2, 2.3, 2.4, ...
And,
Since,
So, irrational number between 2 ans 2.5 are:

Hence, a rational and an irrational number can be 2.1 and $\sqrt{5}$, respectively.

Disclaimer: There are infinite rational and irrational numbers between any two rational numbers.

Answer:

There are infinite number of irrational numbers lying between .

As,
So, the three irrational numbers lying between  are:
1.420420042000..., 1.505005000... and 1.616116111...

Answer:

The two rational numbers between 0.5 and 0.55 are: 0.51 and 0.52

The two irrational numbers between 0.5 and 0.55 are: 0.505005000... and 0.5101100111000...

Disclaimer: There are infinite number of rational and irrational numbers between 0.5 and 0.55.

Answer:

So, the three different irrational numbers are: 0.72020020002..., 0.7515511555111... and 0.808008000...

Disclaimer: There are an infinite number of irrational numbers between two rational numbers.

Answer:

The rational numbers between the numbers 0.2121121112... and 0.2020020002... are:

Disclaimer: There are an infinite number of rational numbers between two irrational numbers.

Answer:

The two irrational numbers between 0.16 and 0.17 are 0.161161116... and 0.1606006000...

Disclaimer: There are an infinite number of irrational numbers between two rational numbers.

Answer:

(i) True

(ii) False
Example:

(iii) True

(iv) False

(v) True

(vi) False
Example:

(vii) False
Real numbers can be divided into rational and irrational numbers.

(viii) True

(ix) True

Answer:

(i)

(ii)

(iii) ${\left(3-\sqrt{3}\right)}^{2}$

(iv) ${\left(\sqrt{5}-\sqrt{3}\right)}^{2}$

(v)

$\left(5+\sqrt{7}\right)\left(2+\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}=5×2+5×\sqrt{5}+\sqrt{7}×2+\sqrt{7}×\sqrt{5}\phantom{\rule{0ex}{0ex}}=10+5\sqrt{5}+2\sqrt{7}+\sqrt{35}$

(vi)

Answer:

$=3×6+3×4\sqrt{2}+\sqrt{3}×6+\sqrt{3}×4\sqrt{2}\phantom{\rule{0ex}{0ex}}=18+12\sqrt{2}+6\sqrt{3}+4\sqrt{6}$

Answer:

(i)

Hence,  is rational.

(ii) ${\left(\sqrt{3}+2\right)}^{2}$

Since, the sum and product of rational numbers and an irrational number is always an irrational.

$⇒$$7+4\sqrt{3}$ is irrational.

Hence, ${\left(\sqrt{3}+2\right)}^{2}$​ is irrational.

(iii) $\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$

$=\frac{2\sqrt{13}}{3\sqrt{13×4}-4\sqrt{13×9}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{13}}{\sqrt{13}\left(3\sqrt{4}-4\sqrt{9}\right)}\phantom{\rule{0ex}{0ex}}=\frac{2}{\left(3×2-4×2\right)}$

Hence, $\frac{2\sqrt{13}}{3\sqrt{52}-4\sqrt{117}}$ is rational.

(iv) $\sqrt{8}+4\sqrt{32}-6\sqrt{2}$

$=2\sqrt{2}+4×4\sqrt{2}-6\sqrt{2}\phantom{\rule{0ex}{0ex}}=2\sqrt{2}+16\sqrt{2}-6\sqrt{2}\phantom{\rule{0ex}{0ex}}=12\sqrt{2}$

Since, the product of a rational number and an irrational number is always an irrational.

Hence, $\sqrt{8}+4\sqrt{32}-6\sqrt{2}$ is rational.

Answer:

(i) As,

Hence, the number of chocolates distributed by Reema is 14.

(ii) The moral values depicted here by Reema is helpfulness and caring.

Disclaimer: The moral values may vary from person to person.

Answer:

(i) $3\sqrt{45}-\sqrt{125}+\sqrt{200}-\sqrt{50}$

$=3\sqrt{9×5}-\sqrt{25×5}+\sqrt{100×2}-\sqrt{25×2}\phantom{\rule{0ex}{0ex}}=3×3\sqrt{5}-5\sqrt{5}+10\sqrt{2}-5\sqrt{2}\phantom{\rule{0ex}{0ex}}=9\sqrt{5}-5\sqrt{5}+5\sqrt{2}\phantom{\rule{0ex}{0ex}}=4\sqrt{5}+5\sqrt{2}$

(ii) $\frac{2\sqrt{30}}{\sqrt{6}}-\frac{3\sqrt{140}}{\sqrt{28}}+\frac{\sqrt{55}}{\sqrt{99}}$

$=\frac{2\sqrt{6×5}}{\sqrt{6}}-\frac{3\sqrt{28×5}}{\sqrt{28}}+\frac{\sqrt{5×11}}{\sqrt{9×11}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}×\sqrt{5}}{\sqrt{6}}-\frac{3\sqrt{28}×\sqrt{5}}{\sqrt{28}}+\frac{\sqrt{5}×\sqrt{11}}{\sqrt{9}×\sqrt{11}}\phantom{\rule{0ex}{0ex}}=2\sqrt{5}-3\sqrt{5}+\frac{\sqrt{5}}{3}$

$=-\sqrt{5}+\frac{\sqrt{5}}{3}\phantom{\rule{0ex}{0ex}}=\frac{-3\sqrt{5}+\sqrt{5}}{3}\phantom{\rule{0ex}{0ex}}=\frac{-2\sqrt{5}}{3}$

(iii) $\sqrt{72}+\sqrt{800}-\sqrt{18}$

$=\sqrt{36×2}+\sqrt{400×2}-\sqrt{9×2}\phantom{\rule{0ex}{0ex}}=6\sqrt{2}+20\sqrt{2}-3\sqrt{2}\phantom{\rule{0ex}{0ex}}=23\sqrt{2}$

Answer:

To represent $\sqrt{5}$ on the number line, follow the following steps of construction:

(i) Mark points 0 and 2 as O and P, respectively.
(ii) At point A, draw AB $\perp$ OA such that AB = 1 units.
(iii) Join OB.
(iv) With O as centre and radius OB, draw an arc intersecting the number line at point P.
Thus, point represents $\sqrt{5}$ on the number line.

Justification:

In right $∆$OAB,

Using Pythagoras theorem,

$\mathrm{OB}=\sqrt{{\mathrm{OA}}^{2}+{\mathrm{AB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{2}^{2}+{1}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+1}\phantom{\rule{0ex}{0ex}}=\sqrt{5}$

Answer:

To represent $\sqrt{3}$
on the number line, follow the following steps of construction:

(i) Mark points 0 and 1 as O and A, respectively.
(ii) At point A, draw AB $\perp$ OA such that AB = 1 units.
(iii) Join OB.
(iv) At point B, draw DB $\perp$ OA such that DB = 1 units.
(v) Join OD.
(vi)
With O as centre and radius OD, draw an arc intersecting the number line at point Q.
Thus, point Q represents $\sqrt{3}$ on the number line.

Justification:

In right $∆$OAB,

Using Pythagoras theorem,

$\mathrm{OB}=\sqrt{{\mathrm{OA}}^{2}+{\mathrm{AB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{1}^{2}+{1}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{1+1}\phantom{\rule{0ex}{0ex}}=\sqrt{2}$

Again, in right
$∆$ODB,

Using Pythagoras theorem,

$\mathrm{OD}=\sqrt{{\mathrm{OB}}^{2}+{\mathrm{DB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\sqrt{2}\right)}^{2}+{1}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{2+1}\phantom{\rule{0ex}{0ex}}=\sqrt{3}$

Answer:

To represent $\sqrt{10}$
on the number line, follow the following steps of construction:

(i) Mark points 0 and 3 as O and B, respectively.
(ii) At point A, draw AB $\perp$ OA such that AB = 1 units.
(iii) Join OA.
(iv) With O as centre and radius OA, draw an arc intersecting the number line at point P.
Thus, point P represents $\sqrt{10}$ on the number line.

Justification:

In right $∆$OAB,

Using Pythagoras theorem,

$\mathrm{OA}=\sqrt{{\mathrm{OB}}^{2}+{\mathrm{AB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{3}^{2}+{1}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+1}\phantom{\rule{0ex}{0ex}}=\sqrt{10}$

Answer:

To represent $\sqrt{8}$
on the number line, follow the following steps of construction:

(i) Mark points 0 and 2 as O and B, respectively.
(ii) At point B, draw AB $\perp$ OA such that AB = 2 units.
(iii) Join OA.
(iv) With O as centre and radius OA, draw an arc intersecting the number line at point P.
Thus, point P represents $\sqrt{8}$ on the number line.

Justification:

In right $∆$OAB,

Using Pythagoras theorem,

$\mathrm{OA}=\sqrt{{\mathrm{OB}}^{2}+{\mathrm{AB}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{2}^{2}+{2}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+4}\phantom{\rule{0ex}{0ex}}=\sqrt{8}$

Answer:

To represent $\sqrt{4.7}$ on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 4.7 units.

(ii) From B, mark a point C on the same given line such that BC = 1 unit.

(iii) Find the mid point of AC and mark it as O.

(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.

(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.

(vi)  With B as centre and radius BD, draw an arc intersecting the given line at point E.

Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents $\sqrt{4.7}$.

Justification:

Here, in semi-circle, radii OA = OC = OD =

And, OB = AB $-$ AO = 4.7 $-$ 2.85 = 1.85 units

In a right angled triangle OBD,

Answer:

To represent $\sqrt{10.5}$ on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 10.5 units.

(ii) From B, mark a point C on the same given line such that BC = 1 unit.

(iii) Find the mid point of AC and mark it as O.

(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.

(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.

(vi)  With B as centre and radius BD, draw an arc intersecting the given line at point E.

Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents $\sqrt{10.5}$.

Justification:

Here, in semi-circle, radii OA = OC = OD =

And, OB = AB $-$ AO = 10.5 $-$ 5.75 = 4.75 units

In a right angled triangle OBD,

Answer:

To represent $\sqrt{7.28}$ on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 7.28 units.

(ii) From B, mark a point C on the same given line such that BC = 1 unit.

(iii) Find the mid point of AC and mark it as O.

(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.

(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.

(vi)  With B as centre and radius BD, draw an arc intersecting the given line at point E.

Thus, let us treat the given line as the number line, with B as 0, C as 1, and so on, then point E represents $\sqrt{7.28}$.

Justification:

Here, in semi-circle, radii OA = OC = OD =

And, OB = AB $-$ AO = 7.28 $-$ 4.14 = 3.14 units

In a right angled triangle OBD,

Answer:

To represent $\left(1+\sqrt{9.5}\right)$ on the number line, follow the following steps of construction:

(i) Mark two points A and B on a given line such that AB = 9.5 units.

(ii) From B, mark a point C on the same given line such that BC = 1 unit.

(iii) Find the mid point of AC and mark it as O.

(iv) With O as centre and radius OC, draw a semi-circle touching the given line at points A and C.

(v) At point B, draw a line perpendicular to AC intersecting the semi-circle at point D.

(vi)  With B as centre and radius BD, draw an arc intersecting the given line at point E.

(vii) From E, mark a point F on the same given line such that EF = 1 unit.

Thus, let us treat the given line as the number line, with B as 0, C as 1, E as $\sqrt{9.5}$ and so on, then point F represents $\left(1+\sqrt{9.5}\right)$.

Justification:

Here, in semi-circle, radii OA = OC = OD =

And, OB = AB $-$ AO = 9.5 $-$ 5.25 = 4.25 units

In a right angled triangle OBD,

Answer:

3 < 3.765 < 4
Divide the gap between 3 and 4 on the number line into 10 equal parts.
Now, 3.7 < 3.765 < 3.8
In order to locate the point 3.765 on the number line, divide the gap between 3.7 and 3.8 into 10 equal parts.
Further, 3.76 < 3.765 < 3.77
So, to locate the point 3.765 on the number line, again divide the gap between 3.76 and 3.77 into 10 equal parts.
Now, the number 3.765 can be located on the number line. This can be shown as follows:

Here, the marked point represents the point 3.765 on the number line.

Answer:

$4.\overline{67}=4.6767$  (Upto 4 decimal places)
4 < 4.6767 < 5
Divide the gap between 4 and 5 on the number line into 10 equal parts.
Now, 4.6 < 4.6767 < 4.7
In order to locate the point 4.6767 on the number line, divide the gap between 4.6 and 4.7 into 10 equal parts.
Further, 4.67 < 4.6767 < 4.68
To locate the point 4.6767 on the number line, again divide the gap between 4.67 and 4.68 into 10 equal parts.
Again, 4.676 < 4.6767 < 4.677
To locate the point 4.6767 on the number line, again divide the gap between 4.676 and 4.677 into 10 equal parts.
Now, the number 4.6767 can be located on the number line. This can be shown as follows:

Here, the marked point represents the point $4.\overline{67}$ on the number line up to 4 decimal places.

Answer:

$\frac{1}{\sqrt{2}+\sqrt{3}}$
$=\frac{1}{\sqrt{3}+\sqrt{2}}×\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{\sqrt{3}-\sqrt{2}}{3-2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}-\sqrt{2}}{1}$
Here, the denominator i.e. 1 is a rational number. Thus, the rationalising factor of the denominator in $\frac{1}{\sqrt{2}+\sqrt{3}}$ is $\sqrt{3}-\sqrt{2}$.

Answer:

(i) $\frac{1}{\sqrt{7}}$
On multiplying the numerator and denominator of the given number by $\sqrt{7}$, we get:

(ii) $\frac{\sqrt{5}}{2\sqrt{3}}$
On multiplying the numerator and denominator of the given number by $\sqrt{3}$, we get:

(iii) $\frac{1}{2+\sqrt{3}}$
On multiplying the numerator and denominator of the given number by $2-\sqrt{3}$, we get:

(iv) $\frac{1}{\sqrt{5}-2}$
On multiplying the numerator and denominator of the given number by $\sqrt{5}+2$, we get:

(v) $\frac{1}{5+3\sqrt{2}}$
On multiplying the numerator and denominator of the given number by $5-3\sqrt{2}$, we get:

(vi) $\frac{1}{\sqrt{7}-\sqrt{6}}$
Multiplying the numerator and denominator by $\sqrt{7}+\sqrt{6}$, we get
$\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}}×\frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}}{{\left(\sqrt{7}\right)}^{2}-{\left(\sqrt{6}\right)}^{2}}$
$=\frac{\sqrt{7}+\sqrt{6}}{7-6}\phantom{\rule{0ex}{0ex}}=\sqrt{7}+\sqrt{6}$

(vii) $\frac{4}{\sqrt{11}-\sqrt{7}}$
Multiplying the numerator and denominator by $\sqrt{11}+\sqrt{7}$, we get
$\frac{4}{\sqrt{11}-\sqrt{7}}=\frac{4}{\sqrt{11}-\sqrt{7}}×\frac{\sqrt{11}+\sqrt{7}}{\sqrt{11}+\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{{\left(\sqrt{11}\right)}^{2}-{\left(\sqrt{7}\right)}^{2}}$
$=\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{11-7}\phantom{\rule{0ex}{0ex}}=\frac{4\left(\sqrt{11}+\sqrt{7}\right)}{4}\phantom{\rule{0ex}{0ex}}=\sqrt{11}+\sqrt{7}$

(viii) $\frac{1+\sqrt{2}}{2-\sqrt{2}}$
Multiplying the numerator and denominator by $2+\sqrt{2}$, we get
$\frac{1+\sqrt{2}}{2-\sqrt{2}}=\frac{1+\sqrt{2}}{2-\sqrt{2}}×\frac{2+\sqrt{2}}{2+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{2+\sqrt{2}+2\sqrt{2}+2}{{\left(2\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{4+3\sqrt{2}}{4-2}\phantom{\rule{0ex}{0ex}}=\frac{4+3\sqrt{2}}{2}$
(ix) $\frac{3-2\sqrt{2}}{3+2\sqrt{2}}$
Multiplying the numerator and denominator by $3-2\sqrt{2}$, we get
$\frac{3-2\sqrt{2}}{3+2\sqrt{2}}=\frac{3-2\sqrt{2}}{3+2\sqrt{2}}×\frac{3-2\sqrt{2}}{3-2\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(3-2\sqrt{2}\right)}^{2}}{{\left(3\right)}^{2}-{\left(2\sqrt{2}\right)}^{2}}$

Answer:

(i)
$\frac{2}{\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{2}{\sqrt{5}}×\frac{\sqrt{5}}{\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{5}}{5}$
$=\frac{2×2.236}{5}\phantom{\rule{0ex}{0ex}}=0.894$
(ii)
$\frac{2-\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2-\sqrt{3}}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{3}-3}{3}$
$=\frac{2×1.732-3}{3}\phantom{\rule{0ex}{0ex}}=0.155$
(iii)
$\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}×\frac{\sqrt{2}}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2}\left(\sqrt{10}-\sqrt{5}\right)}{2}$
$=\frac{1.414×\left(3.162-2.236\right)}{2}\phantom{\rule{0ex}{0ex}}=0.655$

Answer:

(i)
$\frac{\sqrt{2}-1}{\sqrt{2}+1}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2}-1}{\sqrt{2}+1}×\frac{\sqrt{2}-1}{\sqrt{2}-1}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{2}-1\right)}^{2}}{{\left(\sqrt{2}\right)}^{2}-{1}^{2}}$
$=\frac{2+1-2\sqrt{2}}{2-1}\phantom{\rule{0ex}{0ex}}=3-2\sqrt{2}$

(ii)
$\frac{2-\sqrt{5}}{2+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{2-\sqrt{5}}{2+\sqrt{5}}×\frac{2-\sqrt{5}}{2-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(2-\sqrt{5}\right)}^{2}}{{\left(2\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{4+5-4\sqrt{5}}{4-5}\phantom{\rule{0ex}{0ex}}=\frac{9-4\sqrt{5}}{-1}\phantom{\rule{0ex}{0ex}}=-9+4\sqrt{5}$

(iii)
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}×\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^{2}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{3+2+2×\sqrt{3}×\sqrt{2}}{3-2}\phantom{\rule{0ex}{0ex}}=5+2\sqrt{6}$

(iv)
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{5+2\sqrt{3}}{7+4\sqrt{3}}×\frac{7-4\sqrt{3}}{7-4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{35-20\sqrt{3}+14\sqrt{3}-24}{{\left(7\right)}^{2}-{\left(4\sqrt{3}\right)}^{2}}$
$=\frac{11-6\sqrt{3}}{49-48}\phantom{\rule{0ex}{0ex}}=11-6\sqrt{3}$
$\therefore \frac{5+2\sqrt{3}}{7+4\sqrt{3}}=11+\left(-6\right)\sqrt{3}=a+b\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒a=11,b=-6$

Answer:

(i)
$\frac{1}{\sqrt{6}+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{6}+\sqrt{5}}×\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{6}-\sqrt{5}}{{\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{\sqrt{6}-\sqrt{5}}{6-5}\phantom{\rule{0ex}{0ex}}=\sqrt{6}-\sqrt{5}\phantom{\rule{0ex}{0ex}}=2.449-2.236$
$=0.213$
(ii)
$\frac{6}{\sqrt{5}+\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{6}{\sqrt{5}+\sqrt{3}}×\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}$
$=\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{5-3}\phantom{\rule{0ex}{0ex}}=\frac{6\left(\sqrt{5}-\sqrt{3}\right)}{2}\phantom{\rule{0ex}{0ex}}=3\left(\sqrt{5}-\sqrt{3}\right)$
$=3×\left(2.236-1.732\right)\phantom{\rule{0ex}{0ex}}=1.512$
(iii)
$\frac{1}{4\sqrt{3}-3\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{1}{4\sqrt{3}-3\sqrt{5}}×\frac{4\sqrt{3}+3\sqrt{5}}{4\sqrt{3}+3\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{4\sqrt{3}+3\sqrt{5}}{{\left(4\sqrt{3}\right)}^{2}-{\left(3\sqrt{5}\right)}^{2}}$
$=\frac{4\sqrt{3}+3\sqrt{5}}{48-45}\phantom{\rule{0ex}{0ex}}=\frac{4×1.732+3×2.236}{3}\phantom{\rule{0ex}{0ex}}=4.545$
(iv)
$\frac{3+\sqrt{5}}{3-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{3+\sqrt{5}}{3-\sqrt{5}}×\frac{3+\sqrt{5}}{3+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(3+\sqrt{5}\right)}^{2}}{{\left(3\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{9+5+6\sqrt{5}}{9-5}\phantom{\rule{0ex}{0ex}}=\frac{14+6\sqrt{5}}{4}\phantom{\rule{0ex}{0ex}}=\frac{7+3\sqrt{5}}{2}$
$=\frac{7+3×2.236}{2}\phantom{\rule{0ex}{0ex}}=6.854$
(v)
$\frac{1+2\sqrt{3}}{2-\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{1+2\sqrt{3}}{2-\sqrt{3}}×\frac{2+\sqrt{3}}{2+\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2+\sqrt{3}+4\sqrt{3}+6}{{\left(2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}$
$=\frac{8+5\sqrt{3}}{4-3}\phantom{\rule{0ex}{0ex}}=8+5\sqrt{3}\phantom{\rule{0ex}{0ex}}=8+5×1.732$
$=16.660$
(vi)
$\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}×\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{5}+\sqrt{2}\right)}^{2}}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{5+2+2×\sqrt{5}×\sqrt{2}}{5-2}\phantom{\rule{0ex}{0ex}}=\frac{7+2\sqrt{10}}{3}\phantom{\rule{0ex}{0ex}}=\frac{7+2×3.162}{3}$
$=4.441$

Answer:

(i)
$\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{48}+\sqrt{18}}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{3}-5\sqrt{2}}{\sqrt{16×3}+\sqrt{9×2}}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}$
$=\frac{7\sqrt{3}-5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}×\frac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{3}×4\sqrt{3}-7\sqrt{3}×3\sqrt{2}-5\sqrt{2}×4\sqrt{3}+5\sqrt{2}×3\sqrt{2}}{{\left(4\sqrt{3}\right)}^{2}-{\left(3\sqrt{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{84-21\sqrt{6}-20\sqrt{6}+30}{48-18}$
$=\frac{114-41\sqrt{6}}{30}$
(ii)
$\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}×\frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}×3\sqrt{5}+2\sqrt{6}×2\sqrt{6}-\sqrt{5}×3\sqrt{5}-\sqrt{5}×2\sqrt{6}}{{\left(3\sqrt{5}\right)}^{2}-{\left(2\sqrt{6}\right)}^{2}}$
$=\frac{6\sqrt{30}+24-15-2\sqrt{30}}{45-24}\phantom{\rule{0ex}{0ex}}=\frac{9+4\sqrt{30}}{21}$

Answer:

(i)
$\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{4+\sqrt{5}}{4-\sqrt{5}}×\frac{4+\sqrt{5}}{4+\sqrt{5}}+\frac{4-\sqrt{5}}{4+\sqrt{5}}×\frac{4-\sqrt{5}}{4-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(4+\sqrt{5}\right)}^{2}}{{\left(4\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}+\frac{{\left(4-\sqrt{5}\right)}^{2}}{{\left(4\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{16+5+8\sqrt{5}+16+5-8\sqrt{5}}{16-5}\phantom{\rule{0ex}{0ex}}=\frac{42}{11}$
(ii)
$\frac{1}{\sqrt{3}+\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{3}+\sqrt{2}}×\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}-\frac{2}{\sqrt{5}-\sqrt{3}}×\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\frac{3}{\sqrt{2}-\sqrt{5}}×\frac{\sqrt{2}+\sqrt{5}}{\sqrt{2}+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{{\left(\sqrt{2}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{\sqrt{3}-\sqrt{2}}{3-2}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{5-3}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{2-5}\phantom{\rule{0ex}{0ex}}=\sqrt{3}-\sqrt{2}-\frac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}-\frac{3\left(\sqrt{2}+\sqrt{5}\right)}{\left(-3\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{3}-\sqrt{2}-\sqrt{5}-\sqrt{3}+\sqrt{2}+\sqrt{5}$
$=0$
(iii)
$\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}\phantom{\rule{0ex}{0ex}}=\frac{2+\sqrt{3}}{2-\sqrt{3}}×\frac{2+\sqrt{3}}{2+\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}×\frac{2-\sqrt{3}}{2-\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}+1}×\frac{\sqrt{3}-1}{\sqrt{3}-1}\phantom{\rule{0ex}{0ex}}=\frac{{\left(2+\sqrt{3}\right)}^{2}}{{\left(2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}+\frac{{\left(2-\sqrt{3}\right)}^{2}}{{\left(2\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}+\frac{{\left(\sqrt{3}-1\right)}^{2}}{{\left(\sqrt{3}\right)}^{2}-{1}^{2}}$
$=\frac{4+3+4\sqrt{3}}{4-3}+\frac{4+3-4\sqrt{3}}{4-3}+\frac{3+1-2\sqrt{3}}{3-1}\phantom{\rule{0ex}{0ex}}=7+4\sqrt{3}+7-4\sqrt{3}+\frac{4-2\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}=14+2-\sqrt{3}$
$=16-\sqrt{3}$
(iv)
$\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}×\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{6\sqrt{2}}{\sqrt{6}+\sqrt{3}}×\frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{3}}-\frac{8\sqrt{3}}{\sqrt{6}+\sqrt{2}}×\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{6}×\sqrt{3}-2\sqrt{6}×\sqrt{2}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}+\frac{6\sqrt{2}×\sqrt{6}-6\sqrt{2}×\sqrt{3}}{{\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}-\frac{8\sqrt{3}×\sqrt{6}-8\sqrt{3}×\sqrt{2}}{{\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{2}\right)}^{2}}$
$=\frac{2\sqrt{18}-2\sqrt{12}}{3-2}+\frac{6\sqrt{12}-6\sqrt{6}}{6-3}-\frac{8\sqrt{18}-8\sqrt{6}}{6-2}\phantom{\rule{0ex}{0ex}}=2\sqrt{18}-2\sqrt{12}+\frac{6\sqrt{12}-6\sqrt{6}}{3}-\frac{8\sqrt{18}-8\sqrt{6}}{4}\phantom{\rule{0ex}{0ex}}=2\sqrt{18}-2\sqrt{12}+2\sqrt{12}-2\sqrt{6}-2\sqrt{18}+2\sqrt{6}$
$=0$

Answer:

(i)
$\frac{1}{3+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{3}+1}\phantom{\rule{0ex}{0ex}}=\frac{1}{3+\sqrt{7}}×\frac{3-\sqrt{7}}{3-\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{5}}×\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{3}}×\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{3}+1}×\frac{\sqrt{3}-1}{\sqrt{3}-1}\phantom{\rule{0ex}{0ex}}=\frac{3-\sqrt{7}}{{\left(3\right)}^{2}-{\left(\sqrt{7}\right)}^{2}}+\frac{\sqrt{7}-\sqrt{5}}{{\left(\sqrt{7}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}+\frac{\sqrt{5}-\sqrt{3}}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}+\frac{\sqrt{3}-1}{{\left(\sqrt{3}\right)}^{2}-{1}^{2}}$
$=\frac{3-\sqrt{7}}{9-7}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{3}-1}{3-1}\phantom{\rule{0ex}{0ex}}=\frac{3-\sqrt{7}}{2}+\frac{\sqrt{7}-\sqrt{5}}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+\frac{\sqrt{3}-1}{2}\phantom{\rule{0ex}{0ex}}=\frac{3-\sqrt{7}+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-1}{2}$
$=\frac{2}{2}\phantom{\rule{0ex}{0ex}}=1$
(ii)
$\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}\phantom{\rule{0ex}{0ex}}=\frac{1}{1+\sqrt{2}}×\frac{1-\sqrt{2}}{1-\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}×\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}×\frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{5}}×\frac{\sqrt{4}-\sqrt{5}}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{6}}×\frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{7}}×\frac{\sqrt{6}-\sqrt{7}}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{8}}×\frac{\sqrt{7}-\sqrt{8}}{\sqrt{7}-\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}×\frac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}}$
$=\frac{1-\sqrt{2}}{{1}^{2}-{\left(\sqrt{2}\right)}^{2}}+\frac{\sqrt{2}-\sqrt{3}}{{\left(\sqrt{2}\right)}^{2}-{\left(\sqrt{3}\right)}^{2}}+\frac{\sqrt{3}-\sqrt{4}}{{\left(\sqrt{3}\right)}^{2}-{\left(\sqrt{4}\right)}^{2}}+\frac{\sqrt{4}-\sqrt{5}}{{\left(\sqrt{4}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}+\frac{\sqrt{5}-\sqrt{6}}{{\left(\sqrt{5}\right)}^{2}-{\left(\sqrt{6}\right)}^{2}}+\frac{\sqrt{6}-\sqrt{7}}{{\left(\sqrt{6}\right)}^{2}-{\left(\sqrt{7}\right)}^{2}}+\frac{\sqrt{7}-\sqrt{8}}{{\left(\sqrt{7}\right)}^{2}-{\left(\sqrt{8}\right)}^{2}}+\frac{\sqrt{8}-\sqrt{9}}{{\left(\sqrt{8}\right)}^{2}-{\left(\sqrt{9}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+\frac{\sqrt{4}-\sqrt{5}}{4-5}+\frac{\sqrt{5}-\sqrt{6}}{5-6}+\frac{\sqrt{6}-\sqrt{7}}{6-7}+\frac{\sqrt{7}-\sqrt{8}}{7-8}+\frac{\sqrt{8}-\sqrt{9}}{8-9}\phantom{\rule{0ex}{0ex}}=\frac{1-\sqrt{2}}{\left(-1\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(-1\right)}+\frac{\sqrt{3}-\sqrt{4}}{\left(-1\right)}+\frac{\sqrt{4}-\sqrt{5}}{\left(-1\right)}+\frac{\sqrt{5}-\sqrt{6}}{\left(-1\right)}+\frac{\sqrt{6}-\sqrt{7}}{\left(-1\right)}+\frac{\sqrt{7}-\sqrt{8}}{\left(-1\right)}+\frac{\sqrt{8}-\sqrt{9}}{\left(-1\right)}$
$=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}+\sqrt{7}-\sqrt{6}+\sqrt{8}-\sqrt{7}+\sqrt{9}-\sqrt{8}\phantom{\rule{0ex}{0ex}}=3-1\phantom{\rule{0ex}{0ex}}=2$

Answer:

$\frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{7+3\sqrt{5}}{3+\sqrt{5}}×\frac{3-\sqrt{5}}{3-\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}×\frac{3+\sqrt{5}}{3+\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{7\left(3-\sqrt{5}\right)+3\sqrt{5}\left(3-\sqrt{5}\right)}{{3}^{2}-{\left(\sqrt{5}\right)}^{2}}-\frac{7\left(3+\sqrt{5}\right)-3\sqrt{5}\left(3+\sqrt{5}\right)}{{3}^{2}-{\left(\sqrt{5}\right)}^{2}}$
$=\frac{21-7\sqrt{5}+9\sqrt{5}-15}{9-5}-\frac{21+7\sqrt{5}-9\sqrt{5}-15}{9-5}\phantom{\rule{0ex}{0ex}}=\frac{6+2\sqrt{5}}{4}-\frac{6-2\sqrt{5}}{4}$
$=\frac{6+2\sqrt{5}-6+2\sqrt{5}}{4}\phantom{\rule{0ex}{0ex}}=\frac{4\sqrt{5}}{4}\phantom{\rule{0ex}{0ex}}=\sqrt{5}$
$\therefore \frac{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=0+1×\sqrt{5}$
Comparing with the given expression, we get

a = 0 and b = 1

Thus, the values of a and b are 0 and 1, respectively.

Answer:

$\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}+\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}+\sqrt{11}}×\frac{\sqrt{13}-\sqrt{11}}{\sqrt{13}-\sqrt{11}}+\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}×\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}+\sqrt{11}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(\sqrt{13}-\sqrt{11}\right)}^{2}}{{\left(\sqrt{13}\right)}^{2}-{\left(\sqrt{11}\right)}^{2}}+\frac{{\left(\sqrt{13}+\sqrt{11}\right)}^{2}}{{\left(\sqrt{13}\right)}^{2}-{\left(\sqrt{11}\right)}^{2}}$
$=\frac{13+11-2×\sqrt{13}×\sqrt{11}}{13-11}+\frac{13+11+2×\sqrt{13}×\sqrt{11}}{13-11}\phantom{\rule{0ex}{0ex}}=\frac{24-2\sqrt{143}}{2}+\frac{24+2\sqrt{143}}{2}\phantom{\rule{0ex}{0ex}}=\frac{24-2\sqrt{143}+24+2\sqrt{143}}{2}$
$=\frac{48}{2}\phantom{\rule{0ex}{0ex}}=24$

Answer:

$⇒\frac{1}{x}=\frac{1}{3+2\sqrt{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}=\frac{1}{3+2\sqrt{2}}×\frac{3-2\sqrt{2}}{3-2\sqrt{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}=\frac{3-2\sqrt{2}}{{3}^{2}-{\left(2\sqrt{2}\right)}^{2}}$

Adding (1) and (2), we get
$x+\frac{1}{x}=3+2\sqrt{2}+3-2\sqrt{2}=6$, which is a rational number
Thus, $x+\frac{1}{x}$ is rational.

Answer:

Subtracting (2) from (1), we get
$x-\frac{1}{x}=\left(2-\sqrt{3}\right)-\left(2+\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}⇒x-\frac{1}{x}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\phantom{\rule{0ex}{0ex}}⇒{\left(x-\frac{1}{x}\right)}^{3}={\left(-2\sqrt{3}\right)}^{3}=-24\sqrt{3}$
Thus, the value of ${\left(x-\frac{1}{x}\right)}^{3}$ is $-24\sqrt{3}$.

Answer:

Adding (1) and (2), we get
$x+\frac{1}{x}=9-4\sqrt{5}+9+4\sqrt{5}\phantom{\rule{0ex}{0ex}}⇒x+\frac{1}{x}=18$
Squaring on both sides, we get
${\left(x+\frac{1}{x}\right)}^{2}={18}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\frac{1}{{x}^{2}}+2×x×\frac{1}{x}=324\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\frac{1}{{x}^{2}}=324-2=322$
Thus, the value of ${x}^{2}+\frac{1}{{x}^{2}}$ is 322.

Answer:

$⇒\frac{1}{x}=\frac{2}{5-\sqrt{21}}×\frac{5+\sqrt{21}}{5+\sqrt{21}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}=\frac{2\left(5+\sqrt{21}\right)}{{5}^{2}-{\left(\sqrt{21}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}=\frac{2\left(5+\sqrt{21}\right)}{25-21}$

Adding (1) and (2), we get
$x+\frac{1}{x}=\frac{5-\sqrt{21}}{2}+\frac{5+\sqrt{21}}{2}\phantom{\rule{0ex}{0ex}}⇒x+\frac{1}{x}=\frac{5-\sqrt{21}+5+\sqrt{21}}{2}\phantom{\rule{0ex}{0ex}}⇒x+\frac{1}{x}=\frac{10}{2}=5$
Thus, the value of $x+\frac{1}{x}$ is 5.

Answer:

$\therefore \frac{1}{{a}^{2}}=\frac{1}{17-12\sqrt{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{a}^{2}}=\frac{1}{17-12\sqrt{2}}×\frac{17+12\sqrt{2}}{17+12\sqrt{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{a}^{2}}=\frac{17+12\sqrt{2}}{{17}^{2}-{\left(12\sqrt{2}\right)}^{2}}$

Subtracting (2) from (1), we get
${a}^{2}-\frac{1}{{a}^{2}}=\left(17-12\sqrt{2}\right)-\left(17+12\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}⇒{a}^{2}-\frac{1}{{a}^{2}}=17-12\sqrt{2}-17-12\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}-\frac{1}{{a}^{2}}=-24\sqrt{2}$
Thus, the value of ${a}^{2}-\frac{1}{{a}^{2}}$ is $-24\sqrt{2}$.

Answer:

Subtracting (2) from (1), we get

Thus, the value of $x-\frac{1}{x}$ is $4\sqrt{3}$.

Answer:

Adding (1) and (2), we get

Cubing both sides, we get
${\left(x+\frac{1}{x}\right)}^{3}={4}^{3}\phantom{\rule{0ex}{0ex}}⇒{x}^{3}+\frac{1}{{x}^{3}}+3×x×\frac{1}{x}\left(x+\frac{1}{x}\right)=64$
$⇒{x}^{3}+\frac{1}{{x}^{3}}+3×4=64$             [Using (3)]
$⇒{x}^{3}+\frac{1}{{x}^{3}}=64-12=52$
Thus, the value of ${x}^{3}+\frac{1}{{x}^{3}}$ is 52.

Answer:

Disclaimer: The question is incorrect.

$x=\frac{5-\sqrt{3}}{5+\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒x=\frac{5-\sqrt{3}}{5+\sqrt{3}}×\frac{5-\sqrt{3}}{5-\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒x=\frac{{\left(5-\sqrt{3}\right)}^{2}}{{5}^{2}-{\left(\sqrt{3}\right)}^{2}}$
$⇒x=\frac{25+3-10\sqrt{3}}{25-3}\phantom{\rule{0ex}{0ex}}⇒x=\frac{28-10\sqrt{3}}{22}\phantom{\rule{0ex}{0ex}}⇒x=\frac{14-5\sqrt{3}}{11}$

$y=\frac{5+\sqrt{3}}{5-\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒y=\frac{5+\sqrt{3}}{5-\sqrt{3}}×\frac{5+\sqrt{3}}{5+\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒y=\frac{{\left(5+\sqrt{3}\right)}^{2}}{{5}^{2}-{\left(\sqrt{3}\right)}^{2}}$
$⇒y=\frac{25+3+10\sqrt{3}}{25-3}\phantom{\rule{0ex}{0ex}}⇒y=\frac{28+10\sqrt{3}}{22}\phantom{\rule{0ex}{0ex}}⇒y=\frac{14+5\sqrt{3}}{11}$

$\therefore {x}^{2}-{y}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{14-5\sqrt{3}}{11}\right)}^{2}-{\left(\frac{14+5\sqrt{3}}{11}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{196+75-140\sqrt{3}}{121}-\frac{196+75+140\sqrt{3}}{121}\phantom{\rule{0ex}{0ex}}$
$=\frac{271-140\sqrt{3}}{121}-\frac{271+140\sqrt{3}}{121}\phantom{\rule{0ex}{0ex}}=\frac{271-140\sqrt{3}-271-140\sqrt{3}}{121}\phantom{\rule{0ex}{0ex}}=\frac{-280\sqrt{3}}{121}\phantom{\rule{0ex}{0ex}}$
The question is incorrect. Kindly check the question.
The question should have been to show that $x-y=-\frac{10\sqrt{3}}{11}$.
$\therefore x-y\phantom{\rule{0ex}{0ex}}=\frac{14-5\sqrt{3}}{11}-\frac{14+5\sqrt{3}}{11}\phantom{\rule{0ex}{0ex}}=\frac{14-5\sqrt{3}-14-5\sqrt{3}}{11}\phantom{\rule{0ex}{0ex}}=\frac{-10\sqrt{3}}{11}\phantom{\rule{0ex}{0ex}}$

Answer:

According to question,

Now,

Hence, $3{a}^{2}+4ab-3{b}^{2}=4+\frac{56\sqrt{10}}{3}$.

Answer:

According to question,

Now,

Hence, the value of a2 + b2 – 5ab is 93.

Answer:

According to question,

Now,

Hence, the value of p2 + q2 is 47.

Answer:

$\left(\mathrm{i}\right)\frac{1}{\sqrt{7}+\sqrt{6}-\sqrt{13}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\left(\sqrt{7}+\sqrt{6}\right)-\sqrt{13}}×\frac{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}\phantom{\rule{0ex}{0ex}}=\frac{\left(\sqrt{7}+\sqrt{6}\right)+\sqrt{13}}{{\left(\sqrt{7}+\sqrt{6}\right)}^{2}-{\left(\sqrt{13}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{{\left(\sqrt{7}\right)}^{2}+{\left(\sqrt{6}\right)}^{2}+2\left(\sqrt{7}\right)\left(\sqrt{6}\right)-{\left(\sqrt{13}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{7+6+2\sqrt{42}-13}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{7}+\sqrt{6}+\sqrt{13}}{2\sqrt{42}}×\frac{\sqrt{42}}{\sqrt{42}}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{6}+6\sqrt{7}+\left(\sqrt{13}\right)\left(\sqrt{42}\right)}{84}\phantom{\rule{0ex}{0ex}}=\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84}$
Hence, the rationalised form is $\frac{7\sqrt{6}+6\sqrt{7}+\sqrt{546}}{84}$.
$\left(\mathrm{ii}\right)\frac{3}{\sqrt{3}+\sqrt{5}-\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{3}{\left(\sqrt{3}-\sqrt{2}\right)+\sqrt{5}}×\frac{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}}{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}}\phantom{\rule{0ex}{0ex}}=\frac{3\left\{\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right\}}{{\left(\sqrt{3}-\sqrt{2}\right)}^{2}-{\left(\sqrt{5}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{{\left(\sqrt{3}\right)}^{2}+{\left(\sqrt{2}\right)}^{2}-2\left(\sqrt{3}\right)\left(\sqrt{2}\right)-{\left(\sqrt{5}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{3+2-2\sqrt{6}-5}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{-2\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{5}\right]}{-2\sqrt{6}}×\frac{\sqrt{6}}{\sqrt{6}}\phantom{\rule{0ex}{0ex}}=\frac{3\left[3\sqrt{2}-2\sqrt{3}-\sqrt{30}\right]}{-12}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{30}+2\sqrt{3}-3\sqrt{2}}{4}$
Hence, the rationalised form is $\frac{\sqrt{30}+2\sqrt{3}-3\sqrt{2}}{4}$.
$\left(\mathrm{iii}\right)\frac{4}{2+\sqrt{3}+\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{4}{\left(2+\sqrt{3}\right)+\sqrt{7}}×\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\left(2+\sqrt{3}\right)-\sqrt{7}}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{{\left(2+\sqrt{3}\right)}^{2}-{\left(\sqrt{7}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{{\left(2\right)}^{2}+{\left(\sqrt{3}\right)}^{2}+2\left(2\right)\left(\sqrt{3}\right)-{\left(\sqrt{7}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{4+3+4\sqrt{3}-7}\phantom{\rule{0ex}{0ex}}=\frac{4\left[\left(2+\sqrt{3}\right)-\sqrt{7}\right]}{4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{\left(2+\sqrt{3}\right)-\sqrt{7}}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{3}+3-\sqrt{21}}{3}$
Hence, the rationalised form is $\frac{2\sqrt{3}+3-\sqrt{21}}{3}$.

Answer:

Hence, the value of $\frac{1}{\sqrt{3}-\sqrt{2}-1}$ correct to 3 places of decimal is −1.465.

Answer:

Now,

Hence, the value of x3 – 2x2 – 7x + 5 is 3.

Answer:

Hence, $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$ = 5.398 .

Answer:

(i) ${2}^{\frac{2}{3}}×{2}^{\frac{1}{3}}$

(ii) ${2}^{\frac{2}{3}}×{2}^{\frac{1}{5}}$

(iii) ${7}^{\frac{5}{6}}×{7}^{\frac{2}{3}}$

(iv) ${\left(1296\right)}^{\frac{1}{4}}×{\left(1296\right)}^{\frac{1}{2}}$

(i) (ab + ba)–1

(ii) (aa + bb)–1

Answer:

(i) ${\left(\frac{81}{49}\right)}^{-\frac{3}{2}}$

(ii) (14641)0.25

(iii) ${\left(\frac{32}{243}\right)}^{-\frac{4}{5}}$

(iv) ${\left(\frac{7776}{243}\right)}^{-\frac{3}{5}}$

Answer:

(i) $\frac{4}{{\left(216\right)}^{-\frac{2}{3}}}+\frac{1}{{\left(256\right)}^{-\frac{3}{4}}}+\frac{2}{{\left(243\right)}^{-\frac{1}{5}}}$
$\frac{4}{{\left(216\right)}^{-\frac{2}{3}}}+\frac{1}{{\left(256\right)}^{-\frac{3}{4}}}+\frac{2}{{\left(243\right)}^{-\frac{1}{5}}}\phantom{\rule{0ex}{0ex}}=\frac{4}{{\left[{\left(6\right)}^{3}\right]}^{-\frac{2}{3}}}+\frac{1}{{\left[{\left(4\right)}^{4}\right]}^{-\frac{3}{4}}}+\frac{2}{{\left[{\left(3\right)}^{5}\right]}^{-\frac{1}{5}}}\phantom{\rule{0ex}{0ex}}=\frac{4}{{\left(6\right)}^{-2}}+\frac{1}{{\left(4\right)}^{-3}}+\frac{2}{{\left(3\right)}^{-1}}\phantom{\rule{0ex}{0ex}}=4{\left(6\right)}^{2}+{\left(4\right)}^{3}+2\left(3\right)\phantom{\rule{0ex}{0ex}}=144+64+6\phantom{\rule{0ex}{0ex}}=214$

(ii) ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+{\left(\frac{256}{625}\right)}^{-\frac{1}{4}}+{\left(\frac{3}{7}\right)}^{0}$
${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+{\left(\frac{256}{625}\right)}^{-\frac{1}{4}}+{\left(\frac{3}{7}\right)}^{0}\phantom{\rule{0ex}{0ex}}={\left[{\left(\frac{4}{5}\right)}^{3}\right]}^{-\frac{2}{3}}+{\left[{\left(\frac{4}{5}\right)}^{4}\right]}^{-\frac{1}{4}}+1\phantom{\rule{0ex}{0ex}}={\left(\frac{4}{5}\right)}^{-2}+{\left(\frac{4}{5}\right)}^{-1}+1\phantom{\rule{0ex}{0ex}}={\left(\frac{5}{4}\right)}^{2}+\left(\frac{5}{4}\right)+1\phantom{\rule{0ex}{0ex}}=\frac{25}{16}+\frac{5}{4}+1\phantom{\rule{0ex}{0ex}}=\frac{25+20+16}{16}\phantom{\rule{0ex}{0ex}}=\frac{61}{16}$

(iii)

(iv) $\frac{{\left(25\right)}^{\frac{5}{2}}×{\left(729\right)}^{\frac{1}{3}}}{{\left(125\right)}^{\frac{2}{3}}×{\left(27\right)}^{\frac{2}{3}}×{8}^{\frac{4}{3}}}$
$\frac{{\left(25\right)}^{\frac{5}{2}}×{\left(729\right)}^{\frac{1}{3}}}{{\left(125\right)}^{\frac{2}{3}}×{\left(27\right)}^{\frac{2}{3}}×{8}^{\frac{4}{3}}}\phantom{\rule{0ex}{0ex}}=\frac{{\left[{\left(5\right)}^{2}\right]}^{\frac{5}{2}}×{\left[{\left(9\right)}^{3}\right]}^{\frac{1}{3}}}{{\left[{\left(5\right)}^{3}\right]}^{\frac{2}{3}}×{\left[{\left(3\right)}^{3}\right]}^{\frac{2}{3}}×{\left[{\left(2\right)}^{3}\right]}^{\frac{4}{3}}}\phantom{\rule{0ex}{0ex}}=\frac{{\left(5\right)}^{5}×{\left(9\right)}^{1}}{{\left(5\right)}^{2}×{\left(3\right)}^{2}×{\left(2\right)}^{4}}\phantom{\rule{0ex}{0ex}}=\frac{5×5×5×5×5×9}{5×5×3×3×2×2×2×2}\phantom{\rule{0ex}{0ex}}=\frac{125}{16}$

Answer:

(i) ${\left({1}^{3}+{2}^{3}+{3}^{3}\right)}^{\frac{1}{2}}$

(ii) ${\left[5{\left({8}^{\frac{1}{3}}+{27}^{\frac{1}{3}}\right)}^{3}\right]}^{\frac{1}{4}}$

(iii) $\frac{{2}^{0}+{7}^{0}}{{5}^{0}}$

(iv) ${\left[{\left(16\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}}$

Answer:

(i) $\left[{8}^{-\frac{2}{3}}×{2}^{\frac{1}{2}}×{25}^{-\frac{5}{4}}\right]÷\left[{32}^{-\frac{2}{5}}×{125}^{-\frac{5}{6}}\right]=\sqrt{2}$

(ii) ${\left(\frac{64}{125}\right)}^{-\frac{2}{3}}+\frac{1}{{\left(\frac{256}{625}\right)}^{\frac{1}{4}}}+\frac{\sqrt{25}}{\sqrt[3]{64}}=\frac{65}{16}$

(iii) ${\left[7{\left\{{\left(81\right)}^{\frac{1}{4}}+{\left(256\right)}^{\frac{1}{4}}\right\}}^{\frac{1}{4}}\right]}^{4}=16807$

Answer:

Hence, the result in the exponential form is ${x}^{\frac{1}{6}}$.

Answer:

(i) ${\left(\frac{{15}^{\frac{1}{3}}}{{9}^{\frac{1}{4}}}\right)}^{-6}$

(ii) ${\left(\frac{{12}^{\frac{1}{5}}}{{27}^{\frac{1}{5}}}\right)}^{\frac{5}{2}}$

(iii) ${\left(\frac{{15}^{\frac{1}{4}}}{{3}^{\frac{1}{2}}}\right)}^{-2}$

Answer:

Hence, the value of x is 6.

Hence, the value of x is 22.

Hence, the value of x is 5.

Hence, the value of x is 5.

Hence, the value of x is $\frac{5}{4}$.

Answer:

(i) $\sqrt{{x}^{-1}y}·\sqrt{{y}^{-1}z}·\sqrt{{z}^{-1}x}=1$

Hence, $\sqrt{{x}^{-1}y}·\sqrt{{y}^{-1}z}·\sqrt{{z}^{-1}x}=1$.

(ii) ${\left({x}^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left({x}^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}·{\left({x}^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}}=1$

Hence, ${\left({x}^{\frac{1}{a-b}}\right)}^{\frac{1}{a-c}}·{\left({x}^{\frac{1}{b-c}}\right)}^{\frac{1}{b-a}}·{\left({x}^{\frac{1}{c-a}}\right)}^{\frac{1}{c-b}}=1$.

(iii) $\frac{{x}^{a\left(b-c\right)}}{{x}^{b\left(a-c\right)}}÷{\left(\frac{{x}^{b}}{{x}^{a}}\right)}^{c}=1$

Hence, $\frac{{x}^{a\left(b-c\right)}}{{x}^{b\left(a-c\right)}}÷{\left(\frac{{x}^{b}}{{x}^{a}}\right)}^{c}=1$.

(iv)

Hence, .

Answer:

${\left(\frac{{x}^{b}}{{x}^{c}}\right)}^{b+c-a}·{\left(\frac{{x}^{c}}{{x}^{a}}\right)}^{c+a-b}·{\left(\frac{{x}^{a}}{{x}^{b}}\right)}^{a+b-c}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{b+c-a}·{\left({x}^{c-a}\right)}^{c+a-b}·{\left({x}^{a-b}\right)}^{a+b-c}\phantom{\rule{0ex}{0ex}}=\left[{\left({x}^{b-c}\right)}^{b}.{\left({x}^{b-c}\right)}^{c-a}\right]·{\left({x}^{c-a}\right)}^{c+a-b}·\left[{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\right]\phantom{\rule{0ex}{0ex}}=\left[{\left({x}^{b-c}\right)}^{b}.{\left({x}^{b-c}\right)}^{c-a}\right]·{\left({x}^{c+a-b}\right)}^{c-a}·\left[{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\right]\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{b}.\left[{\left({x}^{b-c}\right)}^{c-a}·{\left({x}^{c+a-b}\right)}^{c-a}\right]·{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{b}.\left[{\left({x}^{b-c+c+a-b}\right)}^{c-a}\right]·{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{b}.{\left({x}^{a}\right)}^{c-a}·{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b}\right)}^{b-c}.{\left({x}^{a}\right)}^{c-a}·{\left({x}^{a-b}\right)}^{b-c}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}=\left[{\left({x}^{b}\right)}^{b-c}·{\left({x}^{a-b}\right)}^{b-c}\right].{\left({x}^{a}\right)}^{c-a}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}=\left[{\left({x}^{b+a-b}\right)}^{b-c}\right].{\left({x}^{a}\right)}^{c-a}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{a}\right)}^{b-c}.{\left({x}^{a}\right)}^{c-a}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c}\right)}^{a}.{\left({x}^{c-a}\right)}^{a}.{\left({x}^{a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={\left({x}^{b-c+c-a+a-b}\right)}^{a}\phantom{\rule{0ex}{0ex}}={x}^{0}\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}$

Answer:

$\frac{{9}^{n}×{3}^{2}×{\left({3}^{\frac{-n}{2}}\right)}^{-2}-{\left(27\right)}^{n}}{{3}^{3m}×{2}^{3}}=\frac{1}{27}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left({3}^{2}\right)}^{n}×{3}^{2}×{\left({3}^{-n}\right)}^{-1}-{\left({3}^{3}\right)}^{n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{2n}×{3}^{2}×{3}^{n}-{3}^{3n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{2n+2+n}-{3}^{3n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n+2}-{3}^{3n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n}×{3}^{2}-{3}^{3n}}{{3}^{3m}×{2}^{3}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n}\left(9-1\right)}{{3}^{3m}×8}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n}\left(8\right)}{{3}^{3m}×8}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{{3}^{3n}}{{3}^{3m}}=\frac{1}{{3}^{3}}\phantom{\rule{0ex}{0ex}}⇒{3}^{3n-3m}={3}^{-3}\phantom{\rule{0ex}{0ex}}⇒3n-3m=-3\phantom{\rule{0ex}{0ex}}⇒3\left(n-m\right)=-3\phantom{\rule{0ex}{0ex}}⇒n-m=-1\phantom{\rule{0ex}{0ex}}⇒m-n=1$
Hence, m – n = 1.

Answer:

Since, the sum and product of a rational and an irrational is always irrational.

So, $1+\sqrt{3}$ and $2\sqrt{3}$ are irrational numbers.

Also, π is an irrational number.

And, 0 is an integer.

So, 0 is a rational number.

Hence, the correct option is (d).

Answer:

Since, –4.3 < –3.4 < –3 < 0 < 1.101100110001... < 3

But 1.101100110001... is an irrational number

So, the rational number between –3 and 3 is 0.

Hence, the correct option is (a).

Answer:

We have,

And,

Also,

Since, $\frac{1}{6}<\frac{2}{6}<\frac{3}{6}\left(\frac{1}{2}\right)<\frac{4}{6}\left(=\frac{2}{3}\right)<\frac{5}{6}<\frac{7}{6}<\frac{8}{6}\left(=\frac{4}{3}\right)<\frac{10}{6}\left(=\frac{5}{3}\right)<\frac{12}{6}\left(=\frac{2}{1}\right)$

So, the two rational numbers between  are .

Hence, the correct opion is (c).

Answer:

As, all rational numbers, all natural numbers and all irrational numbers can be represented on a nuumber line in an unique way.

So, every point on a number line represents a unique number.

Hence, the correct option is (d).

Answer:

(c) $\sqrt{225}$

Because 225 is a square of 15, i.e., $\sqrt{225}$ = 15, and it can be expressed in the $\frac{p}{q}$ form, it is a rational number.

Answer:

(d) a real number

Every rational number is a real number, as every rational number can be easily expressed on the real number line.

Answer:

(c) are infinitely many rational numbers

Because the range between any two rational numbers can be easily divided into any number of divisions, there can be an infinite number of rational numbers between any two rational numbers.

Answer:

(b) either terminating or repeating

As per the definition of rational numbers, they are either repeating or terminating decimals.

Answer:

(d) neither terminating nor repeating

As per the definition of irrational numbers, these are neither terminating nor repeating decimals.

Answer:

As, any number which have a terminating or non-terminating recurring decimal expansion is a rational number.

So, 0.5030030003... which is non-termintaing non-recurring decimal expansion is not a rational number.

Hence, the correct option is (d).

Answer:

(d) 3.141141114...

Because 3.141141114... is neither a repeating decimal nor a terminating decimal, it is an irrational number.

Answer:

Since, $\frac{7}{19}=\frac{7×3}{19×3}=\frac{21}{57}$

Hence, the correct option is (d).

Answer:

We have,

And,

Since, $-\frac{40}{60}\left(=-\frac{2}{3}\right)<-\frac{21}{60}\left(=-\frac{7}{20}\right)<-\frac{18}{60}\left(=-\frac{3}{10}\right)<-\frac{15}{60}\left(=-\frac{1}{4}\right)<-\frac{12}{60}\left(=-\frac{1}{5}\right)<\frac{18}{60}\left(=\frac{3}{10}\right)$

So, the rational number which does not lie between  is $\frac{3}{10}$.

Hence, the correct option is (b).

Answer:

Since, π has a non-terminating non-recurring decimal expansion.

So, π is an irrational number.

Hence, the correct option is (c).

Answer:

(c) a non-terminating and non-repeating decimal

Because $\sqrt{2}$ is an irrational number, its decimal expansion is non-terminating and non-repeating.

Answer:

Since, $\sqrt{225}$ = 15, which is an integer,

0.3799 is a number with terminating decimal expansion, and

$7.\overline{478}$ is a number with non-terminating recurring decimal expansion

Also, 23 is a prime number.

So, $\sqrt{23}$ is an irrational number.

Hence, the correct option is (a).

Answer:

So, there are 6 digits in the repeating block of digits in the decimal expansion of $\frac{17}{7}$.

Hence, the correct option is (b).

Answer:

Since,

$\sqrt{\frac{4}{9}}=\frac{2}{3}$, which is a rational number,

$\frac{\sqrt{1250}}{\sqrt{8}}=\sqrt{\frac{1250}{8}}=\sqrt{\frac{625}{4}}=\frac{25}{2}$, which is a rational number,

$\sqrt{8}=2\sqrt{2}$, which is an irrational number, and

$\frac{\sqrt{24}}{\sqrt{6}}=\sqrt{\frac{24}{6}}=\sqrt{4}=2$, which is a rational number

Hence, the correct option is (c).

Answer:

(d) sometimes rational and sometimes irrational

For example:
$\sqrt{2}$ is an irrational number, when it is multiplied with itself  it results into 2, which is a rational number.
$\sqrt{2}$  when multiplied with $\sqrt{3}$, which is also an irrational number, results into $\sqrt{6}$, which is an irrational number.

Answer:

(d) Every real number is either rational or irrational.

Because a real number can be further categorised into either a rational number or an irrational number, every real number is either rational or irrational.

Answer:

(d) $\mathrm{\pi }$ is irrational and $\frac{22}{7}$ is rational.
Because the value of $\mathrm{\pi }$ is neither repeating nor terminating, it is an irrational number. $\frac{22}{7}$, on the other hand, is of the form $\frac{p}{q}$, so it is a rational number.

Answer:

Since, $\frac{\left(\sqrt{2}+\sqrt{3}\right)}{2}$ and $\sqrt{6}$ are irrational numbers,

And,

So, the rational number lying between  is 1.6 .

Hence, the correct option is (c).

Answer:

Since, a number whose decimal expansion is terminating or non-terminating recurring is rational number.

So, 0.853853853... is a rational number.

Hence, the correct option is (d).

Answer:

Since, the product of a non-zero rational number with an irrational number is always an irrational number.

Hence, the correct option is (a).

Answer:

Let
Multiplying both sides by 10, we get

Subtracting (1) from (2), we get
$10x-x=2.\overline{)2}-0.\overline{)2}\phantom{\rule{0ex}{0ex}}⇒9x=2\phantom{\rule{0ex}{0ex}}⇒x=\frac{2}{9}\phantom{\rule{0ex}{0ex}}\therefore 0.\overline{)2}=\frac{2}{9}$
Hence, the correct answer is option (b).

Answer:

(c) $\frac{5}{3}$
Let x = 1.6666666...       ...(i)
Multiplying by 10 on both sides, we get:
10x = 16.6666666...       ...(ii)
Subtracting (i) from (ii), we get:
9x = 15
$⇒$x =

Answer:

(b) $\frac{6}{11}$
Let x = 0.545454...               ...(i)
Multiplying both sides by 100, we get:
100x = 54.5454545...           ...(ii)
Subtracting (i) from (ii), we get:
99x = 540
$⇒$x$\frac{54}{99}$ = $\frac{6}{11}$

Answer:

(c) $\frac{29}{90}$
Let x = 0.3222222222...          ...(i)
Multiplying by 10 on both sides, we get:
10x = 3.222222222...              ...(ii)
Again, multiplying by 10 on both sides, we get:
100x = 32.222222222...          ...(iii)
On subtracting (ii) from (iii), we get:
90x = 29
x = $\frac{29}{90}$

Answer:

(d) none of these
Let x = 0.12333333333...         ...(i)
Multiplying by 100 on both sides, we get:
100x = 12.33333333...             ...(ii)
Multiplying by 10 on both sides, we get:
1000x = 123.33333333...         ...(iii)
Subtracting (ii) from (iii), we get:
900x = 111

$⇒$x = $\frac{111}{900}$

Answer:

(c) $\sqrt{5×6}$

An irrational number between a and b is given as $\sqrt{ab}$.

Answer:

(d) 61/4
An irrational number between

Answer:

(c) $\sqrt{\frac{1}{7}×\frac{2}{7}}$

An irrational number between a and b is given as $\sqrt{ab}$.

Answer:

Let
Multiplying both sides by 10, we get

Subtracting (1) from (2), we get
$10x-x=3.\overline{)3}-0.\overline{)3}\phantom{\rule{0ex}{0ex}}⇒9x=3\phantom{\rule{0ex}{0ex}}⇒x=\frac{3}{9}\phantom{\rule{0ex}{0ex}}\therefore 0.\overline{)3}=\frac{3}{9}$
Let
Multiplying both sides by 10, we get

Subtracting (3) from (4), we get
$10y-y=4.\overline{)4}-0.\overline{)4}\phantom{\rule{0ex}{0ex}}⇒9y=4\phantom{\rule{0ex}{0ex}}⇒y=\frac{4}{9}\phantom{\rule{0ex}{0ex}}\therefore 0.\overline{)4}=\frac{4}{9}$
Sum of $0.\overline{3}$ and $0.\overline{4}$ = $0.\overline{)3}+0.\overline{)4}=\frac{3}{9}+\frac{4}{9}=\frac{7}{9}$
Hence, the correct answer is option (b).

Answer:

Let
Multiplying both sides by 100, we get

Subtracting (1) from (2), we get
$100x-x=245.\overline{)45}-2.\overline{)45}\phantom{\rule{0ex}{0ex}}⇒99x=245-2=243\phantom{\rule{0ex}{0ex}}⇒x=\frac{243}{99}\phantom{\rule{0ex}{0ex}}\therefore 2.\overline{)45}=\frac{243}{99}$
Let
Multiplying both sides by 100, we get

Subtracting (3) from (4), we get
$100y-y=36.\overline{)36}-0.\overline{)36}\phantom{\rule{0ex}{0ex}}⇒99y=36\phantom{\rule{0ex}{0ex}}⇒y=\frac{36}{99}\phantom{\rule{0ex}{0ex}}\therefore 0.\overline{)36}=\frac{36}{99}$
So, $2.\overline{)45}+0.\overline{)36}=\frac{243}{99}+\frac{36}{99}=\frac{243+36}{99}=\frac{279}{99}=\frac{31}{11}$
Hence, the correct answer is option (c).

Answer:

Hence, the correct answer is option (b).

Answer:

$\left(-2-\sqrt{3}\right)\left(-2+\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=-\left(2+\sqrt{3}\right)×\left[-\left(2-\sqrt{3}\right)\right]\phantom{\rule{0ex}{0ex}}=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)$

Thus, the given expression when simplified is positive and rational.

Hence, the correct answer is option (b).

Answer:

$\left(6+\sqrt{27}\right)-\left(3+\sqrt{3}\right)+\left(1-2\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=6+\sqrt{3×3×3}-\left(3+\sqrt{3}\right)+\left(1-2\sqrt{3}\right)\phantom{\rule{0ex}{0ex}}=6+3\sqrt{3}-3-\sqrt{3}+1-2\sqrt{3}\phantom{\rule{0ex}{0ex}}=4$
Thus, the given expression when simplified is positive and rational.

Hence, the correct answer is option (b).

Answer:

Hence, the correct answer is option (c).

Answer:

$\sqrt{20}×\sqrt{5}\phantom{\rule{0ex}{0ex}}=\sqrt{2×2×5}×\sqrt{5}\phantom{\rule{0ex}{0ex}}=2\sqrt{5}×\sqrt{5}$
$=2×5\phantom{\rule{0ex}{0ex}}=10$
Hence, the correct answer is option (a).

Answer:

$\frac{4\sqrt{12}}{12\sqrt{27}}\phantom{\rule{0ex}{0ex}}=\frac{4\sqrt{2×2×3}}{12\sqrt{3×3×3}}\phantom{\rule{0ex}{0ex}}=\frac{4×2\sqrt{3}}{12×3\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{2}{9}$
Hence, the correct answer is option (b).

Answer:

$=5×\sqrt{2×3}\phantom{\rule{0ex}{0ex}}=5\sqrt{6}$
Hence, the correct answer is option (ii).

Answer:

$=\frac{4\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}\phantom{\rule{0ex}{0ex}}=\frac{4}{2}\phantom{\rule{0ex}{0ex}}=2$
Hence, the correct answer is option (b).

Answer:

Hence, the correct answer is option (c).

Answer:

Hence, the correct answer is option (b).

Answer:

Hence, the correct answer is option (d).

Answer:

$\therefore$ The value of $\sqrt[4]{{\left(64\right)}^{-2}}$ is $\frac{1}{8}$.

Hence, the correct option is (a).

Answer:

$\therefore$ The value of $\frac{{2}^{0}+{7}^{0}}{{5}^{0}}$ is 2.

Hence, the correct option is (b).

Answer:

$\therefore$ The value of ${\left(243\right)}^{\frac{1}{5}}$ is 3.

Hence, the correct option is (a).

Answer:

Hence, the correct option is (c).

Answer:

$\therefore$Simplified value of ${\left(16\right)}^{-\frac{1}{4}}×\sqrt[4]{16}$ is 1.

Hence, the correct option is (b).

Answer:

$\therefore$The value of  $\sqrt[4]{\sqrt[3]{{2}^{2}}}$ is ${2}^{\frac{1}{6}}$.

Hence, the correct option is (c).

Answer:

$\therefore$Simplified value of ${\left(25\right)}^{\frac{1}{3}}×{5}^{\frac{1}{3}}$ is 5.

Hence, the correct option is (d).

Answer:

$\therefore$The value of ${\left[{\left(81\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}}$ is 3.

Hence, the correct option is (a).

Answer:

$\therefore$ x can be $\sqrt[4]{2}$.

Hence, the correct option is (d).

Answer:

Hence, the correct option is (b).

Answer:

Hence, the correct option is (b).

Answer:

Hence, the correct option is (b).

Answer:

Hence, the correct option is (c).

Answer:

Hence, the correct option is (a).

Answer:

Hence, the correct answer is option (d).

Answer:

Hence, the correct answer is option (d).

Answer:

$\frac{{5}^{n+2}-6×{5}^{n+1}}{13×{5}^{n}-2×{5}^{n+1}}\phantom{\rule{0ex}{0ex}}=\frac{{5}^{n}×{5}^{2}-6×{5}^{n}×5}{13×{5}^{n}-2×{5}^{n}×5}\phantom{\rule{0ex}{0ex}}=\frac{{5}^{n}×5\left[5-6\right]}{{5}^{n}\left[13-2×5\right]}\phantom{\rule{0ex}{0ex}}=\frac{-5}{3}$
Hence, the correct answer is option (b).

Answer:

$\sqrt[3]{500}=\sqrt[3]{5×5×2×5×2}=\sqrt[3]{{5}^{3}×{2}^{2}}=5\sqrt[3]{4}$
So, the simplest rationalisation factor of $\sqrt[3]{500}$ is $\sqrt[3]{2}$.
Hence, the correct answer is option (d).

Answer:

Simplest rationalisation ractor of $\left(2\sqrt{2}-\sqrt{3}\right)$ is $2\sqrt{2}+\sqrt{3}$.
Hence, the correct answer is option (b).

Answer:

Rationalisation factor of $\frac{1}{2\sqrt{3}-\sqrt{5}}$ will be $2\sqrt{3}+\sqrt{5}=\sqrt{4×3}+\sqrt{5}=\sqrt{12}+\sqrt{5}$.
Hence, the correct answer is option (d).

Answer:

$\frac{1}{\sqrt{5}+\sqrt{2}}$
Rationalisation of denominator gives
$\frac{1}{\sqrt{5}+\sqrt{2}}×\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{5-2}=\frac{\sqrt{5}-\sqrt{2}}{3}$
Hence, the correct answer is option (d).

Answer:

$x=2+\sqrt{3}\phantom{\rule{0ex}{0ex}}\frac{1}{x}=\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}×\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$
$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4$
Hence, the correct answer is option (c).

Answer:

$\frac{1}{\left(3+2\sqrt{2}\right)}=\frac{1}{\left(3+2\sqrt{2}\right)}×\frac{\left(3-2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)}=\frac{\left(3-2\sqrt{2}\right)}{9-8}=\left(3-2\sqrt{2}\right)$
Hence, the correct answer is option (c).

Answer:

Given: $x=\left(7+4\sqrt{3}\right)$
$\frac{1}{x}=\frac{1}{7+4\sqrt{3}}=\frac{1}{7+4\sqrt{3}}×\frac{7-4\sqrt{3}}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}}{49-48}=7-4\sqrt{3}$
$\left(x+\frac{1}{x}\right)=7+4\sqrt{3}+\left(7-4\sqrt{3}\right)=14$
Hence, the correct answer is option (b).

Answer:

Hence, the correct answer is option (c).

Answer:

$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}×\frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}$
Given that

Hence, the correct answer is option (b).

Answer:

$3-2\sqrt{2}=2+1-2×\sqrt{2}×1\phantom{\rule{0ex}{0ex}}={\left(\sqrt{2}\right)}^{2}+{1}^{2}-2×\sqrt{2}×1$
This is of the form
${a}^{2}+{b}^{2}-2ab={\left(a-b\right)}^{2}$

Hence, the correct answer is option (d).

Answer:

$5+2\sqrt{6}=2+3+2×\sqrt{3}×\sqrt{2}\phantom{\rule{0ex}{0ex}}={\left(\sqrt{2}\right)}^{2}+{\left(\sqrt{3}\right)}^{2}+2×\sqrt{3}×\sqrt{2}$
This is in the form
${a}^{2}+{b}^{2}+2ab={\left(a+b\right)}^{2}$
So, we have
${\left(\sqrt{2}\right)}^{2}+{\left(\sqrt{3}\right)}^{2}+2×\sqrt{3}×\sqrt{2}={\left(\sqrt{2}+\sqrt{3}\right)}^{2}$
Thus, $\sqrt{5+2\sqrt{6}}=\sqrt{{\left(\sqrt{2}+\sqrt{3}\right)}^{2}}=\sqrt{2}+\sqrt{3}$
Hence, the correct answer is option (c).

Answer:

$\sqrt{\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)}}=\sqrt{\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)}×\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)}}=\sqrt{\frac{{\left(\sqrt{2}-1\right)}^{2}}{2-1}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{2+1-2\sqrt{2}}{1}}\phantom{\rule{0ex}{0ex}}=\sqrt{3-2\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{3-2×1.414}\phantom{\rule{0ex}{0ex}}=\sqrt{3-2.828}\phantom{\rule{0ex}{0ex}}=\sqrt{0.172}\phantom{\rule{0ex}{0ex}}=0.414$
Hence, the correct answer is option (c).

Answer:

Given: $x=3+\sqrt{8}$
$\frac{1}{x}=\frac{1}{3+\sqrt{8}}=\frac{1}{3+\sqrt{8}}×\frac{3-\sqrt{8}}{3-\sqrt{8}}=\frac{3-\sqrt{8}}{9-8}=\frac{3-\sqrt{8}}{1}=3-\sqrt{8}$
$x+\frac{1}{x}=\left(3+\sqrt{8}\right)+\left(3-\sqrt{8}\right)=6$
${\left(x+\frac{1}{x}\right)}^{2}={x}^{2}+\frac{1}{{x}^{2}}+2×x×\frac{1}{x}={x}^{2}+\frac{1}{{x}^{2}}+2\phantom{\rule{0ex}{0ex}}⇒{6}^{2}={x}^{2}+\frac{1}{{x}^{2}}+2\phantom{\rule{0ex}{0ex}}⇒36={x}^{2}+\frac{1}{{x}^{2}}+2\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+\frac{1}{{x}^{2}}=36-2=34$
Hence, the correct answer is option (a).

Answer:

(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

So, Assertion and Reason are correct (property of rational numbers). Also, Reason is the correct explanation of Assertion.

Answer:

(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

3 3

Answer:

(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.

It is known that e and $\mathrm{\pi }$ are irrational numbers, but Reason is not the correct explanation.

Answer:

(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.

Answer:

(a) Because it is a non-terminating and repeating decimal, it is a rational number.

(b) $\mathrm{\pi }$ is an irrational number.

(c) $\frac{1}{7}=.142857142857$...
Hence, its period is 6.

(d)

Answer:

(a)
${\left({\left(81\right)}^{-2}\right)}^{\frac{1}{4}}\phantom{\rule{0ex}{0ex}}={\left({\left(9\right)}^{-4}\right)}^{\frac{1}{4}}={\left(9\right)}^{-4×\frac{1}{4}}={\left(9\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{9}$

(b)

(c)

(d)

${\left(\frac{3}{2}\right)}^{4×\frac{-3}{4}}×{\left(\frac{4}{3}\right)}^{3×\frac{-1}{3}}\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{2}\right)}^{-3}×{\left(\frac{4}{3}\right)}^{-1}\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{2}\right)}^{-3}×\frac{3}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\left(\frac{{3}^{-3}}{{2}^{-3}}\right)×\frac{3}{{2}^{2}}\phantom{\rule{0ex}{0ex}}=\left(\frac{{3}^{-3}×3}{{2}^{-3}×{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{{3}^{-2}}{{2}^{-1}}\phantom{\rule{0ex}{0ex}}=\frac{2}{9}$

Answer:

Sum of a rational number and an irrational number is an irrational number.
Example: 4 + $\sqrt{5}$ represents sum of rational and an irrational number where 4 is rational and $\sqrt{5}$ is irrational.

Answer:

$\frac{665}{625}=\frac{5×19×7}{{5}^{4}}=\frac{19×7}{{5}^{3}}=\frac{19×7×{2}^{3}}{{5}^{3}×{2}^{3}}=\frac{1064}{1000}=1.064$
So, $\frac{665}{625}$ will terminate after 3 decimal places.

Answer:

${\left(1296\right)}^{0.17}×{\left(1296\right)}^{0.08}\phantom{\rule{0ex}{0ex}}={\left(1296\right)}^{0.17+0.08}\phantom{\rule{0ex}{0ex}}={\left(1296\right)}^{0.25}\phantom{\rule{0ex}{0ex}}={\left(1296\right)}^{\frac{1}{4}}\phantom{\rule{0ex}{0ex}}=\sqrt[4]{1296}\phantom{\rule{0ex}{0ex}}=6$

Answer:

$6\sqrt{36}+5\sqrt{12}\phantom{\rule{0ex}{0ex}}=6×6+5\sqrt{4×3}\phantom{\rule{0ex}{0ex}}=36+10\sqrt{3}$

Answer:

A number which is non terminating and non recurring is known as irrational number.

There are infinitely many irrational numbers between 5 and 6.

One of the example is 5.40430045000460000....

Answer:

Hence, the value of $\frac{21\sqrt{12}}{10\sqrt{27}}$ is $\frac{7}{5}$.

Answer:

Hence, the rationalised form is $\sqrt{3}-\sqrt{2}$.

Answer:

${\left(\frac{2}{5}\right)}^{2x-2}=\frac{32}{3125}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{2}{5}\right)}^{2x-2}=\frac{{2}^{5}}{{5}^{5}}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{2}{5}\right)}^{2x-2}={\left(\frac{2}{5}\right)}^{5}\phantom{\rule{0ex}{0ex}}⇒2x-2=5\phantom{\rule{0ex}{0ex}}⇒2x=5+2\phantom{\rule{0ex}{0ex}}⇒x=\frac{7}{2}$
Hence, $x=\frac{7}{2}$.

Answer:

Hence, ${\left(32\right)}^{\frac{1}{5}}+{\left(-7\right)}^{0}+{\left(64\right)}^{\frac{1}{2}}$ = 11.

Answer:

Hence, ${\left(\frac{81}{49}\right)}^{\frac{-3}{2}}=\frac{343}{729}$.

Answer:

For = 1and b = 2,
${\left({a}^{b}+{b}^{a}\right)}^{-1}\phantom{\rule{0ex}{0ex}}={\left({1}^{2}+{2}^{1}\right)}^{-1}\phantom{\rule{0ex}{0ex}}={\left(1+2\right)}^{-1}$

Thus, the value of (ab + ba)–1 when a = 1 and b = 2 is $\frac{1}{3}$.

Answer:

Let the two irrational numbers be $2+\sqrt{3}$ and $2-\sqrt{3}$.
Sum of these irrational numbers $=\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)=4$, which is rational
Product of these irrational numbers $=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)={2}^{2}-{\left(\sqrt{3}\right)}^{2}=4-3=1$, which is rational

Answer:

Yes, the product of a rational and an irrational number is always an irrational number.
Example:
2 is a rational number and $\sqrt{3}$ is an irrational number.
Now, $2×\sqrt{3}=2\sqrt{3}$, which is an irrational number.

Answer:

The cube roots of natural numbers which are not perfect cubes are all irrational numbers.
Let $x=\sqrt[3]{2}={2}^{\frac{1}{3}}$.
Now,
${x}^{2}={\left({2}^{\frac{1}{3}}\right)}^{2}={2}^{\frac{2}{3}}={\left({2}^{2}\right)}^{\frac{1}{3}}={4}^{\frac{1}{3}}$, which is an irrational number
Also,
${x}^{3}={\left({2}^{\frac{1}{3}}\right)}^{3}={2}^{3×\frac{1}{3}}=2$, which is a rational number

Answer:

The reciprocal of $\left(2+\sqrt{3}\right)$

$=\frac{\left(2-\sqrt{3}\right)}{4-3}\phantom{\rule{0ex}{0ex}}=\frac{\left(2-\sqrt{3}\right)}{1}\phantom{\rule{0ex}{0ex}}=\left(2-\sqrt{3}\right)$

Answer:

The value of $\frac{1}{\sqrt{10}}=\frac{1}{\sqrt{10}}×\frac{\sqrt{10}}{\sqrt{10}}=\frac{\sqrt{10}}{10}=\frac{3.162}{10}=0.3162$

Answer:

We have,

${10}^{x}=64$

Taking square root from both sides, we get

$\sqrt{{10}^{x}}=\sqrt{64}\phantom{\rule{0ex}{0ex}}⇒{\left({10}^{x}\right)}^{\frac{1}{2}}=8\phantom{\rule{0ex}{0ex}}⇒{10}^{\left(\frac{x}{2}\right)}=8$

Multiplying both sides by 10, we get

Answer:

$\frac{{2}^{n}+{2}^{n-1}}{{2}^{n+1}-{2}^{n}}\phantom{\rule{0ex}{0ex}}=\frac{{2}^{n}\left(1+{2}^{-1}\right)}{{2}^{n}\left(2-1\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(1+\frac{1}{2}\right)}{1}\phantom{\rule{0ex}{0ex}}=\frac{2+1}{2}\phantom{\rule{0ex}{0ex}}=\frac{3}{2}$

Answer:

${\left[{\left\{{\left(256\right)}^{-\frac{1}{2}}\right\}}^{\frac{1}{4}}\right]}^{2}\phantom{\rule{0ex}{0ex}}={\left\{{\left(256\right)}^{-\frac{1}{2}}\right\}}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}={\left(256\right)}^{-\frac{1}{4}}\phantom{\rule{0ex}{0ex}}={\left({4}^{4}\right)}^{-\frac{1}{4}}\phantom{\rule{0ex}{0ex}}={4}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}$

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