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Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

________

$5ab$

Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

_____

Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

___________

Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise, we get:

Page No 84:

On arranging the terms of the given expressions in the descending powers of $x$ and adding column-wise:

Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and adding column-wise:

Page No 84:

On arranging the terms of the given expressions in the descending powers of $x$ and adding column-wise:

Page No 84:

On arranging the terms of the given expressions in the descending powers of $x$ and subtracting:

Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

Page No 84:

On arranging the terms of the given expressions in the descending powers of $x$ and subtracting column-wise:

Page No 84:

Arranging the terms of the given expressions in the descending powers of $x$ and subtracting column-wise:

Page No 84:

Writing the terms of the given expressions (in the same order) in the form of rows with like terms below each other and subtracting column-wise:

Page No 84:

Let the required number be $x$.
$\left(3{a}^{2}-6ab-3{b}^{2}-1\right)-x=4{a}^{2}-7ab-4{b}^{2}+1$
$\left(3{a}^{2}-6ab-3{b}^{2}-1\right)-\left(4{a}^{2}-7ab-4{b}^{2}+1\right)=x\phantom{\rule{0ex}{0ex}}$

∴ Required number = $-{a}^{2}+ab+{b}^{2}-2$

Page No 84:

Sides of the rectangle are $l$ and $b$.
$l=5{x}^{2}-3{y}^{2}\phantom{\rule{0ex}{0ex}}b={x}^{2}+2xy$
Perimeter of the rectangle is $\left(2l+2b\right)$.

Page No 84:

Let be the three sides of the triangle.

∴ Perimeter of the triangle =$\left(a+b+c\right)$

Given perimeter of the triangle = $6{p}^{2}-4p+9$
One side ($a$)  = ${p}^{2}-2p+1$
Other side ($b$) = $3{p}^{2}-5p+3$
Perimeter = $\left(a+b+c\right)$

Thus, the third side is $2{p}^{2}+3p+5$.

Page No 87:

By horizontal method:
$\left(5x+7\right)×\left(3x+4\right)\phantom{\rule{0ex}{0ex}}=5x\left(3x+4\right)+7\left(3x+4\right)\phantom{\rule{0ex}{0ex}}=15{x}^{2}+20x+21x+28\phantom{\rule{0ex}{0ex}}=15{x}^{2}+41x+28$

Page No 87:

By horizontal method:

$\left(4x+9\right)×\left(x-6\right)\phantom{\rule{0ex}{0ex}}=4x\left(x-6\right)+9\left(x-6\right)\phantom{\rule{0ex}{0ex}}=4{x}^{2}-24x+9x-54\phantom{\rule{0ex}{0ex}}=4{x}^{2}-15x-54$

Page No 87:

By horizontal method:

$\left(2x+5\right)×\left(4x-3\right)\phantom{\rule{0ex}{0ex}}=2x\left(4x-3\right)+5\left(4x-3\right)\phantom{\rule{0ex}{0ex}}=8{x}^{2}-6x+20x-15\phantom{\rule{0ex}{0ex}}=8{x}^{2}+14x-15$

Page No 87:

By horizontal method:

$\left(3y-8\right)×\left(5y-1\right)\phantom{\rule{0ex}{0ex}}=3y\left(5y-1\right)-8\left(5y-1\right)\phantom{\rule{0ex}{0ex}}=15{y}^{2}-3y-40y+8\phantom{\rule{0ex}{0ex}}=15{y}^{2}-43y+8$

Page No 87:

By horizontal method:

$\left(7x+2y\right)×\left(x+4y\right)\phantom{\rule{0ex}{0ex}}=7x\left(x+4y\right)+2y\left(x+4y\right)\phantom{\rule{0ex}{0ex}}=7{x}^{2}+28xy+2xy+8{y}^{2}\phantom{\rule{0ex}{0ex}}=7{x}^{2}+30xy+8{y}^{2}$

Page No 87:

By horizontal method:

$\left(9x+5y\right)×\left(4x+3y\right)\phantom{\rule{0ex}{0ex}}9x\left(4x+3y\right)+5y\left(4x+3y\right)\phantom{\rule{0ex}{0ex}}=36{x}^{2}+27xy+20xy+15{y}^{2}\phantom{\rule{0ex}{0ex}}=36{x}^{2}+47xy+15{y}^{2}$

Page No 87:

By horizontal method:

$\left(3m-4n\right)×\left(2m-3n\right)\phantom{\rule{0ex}{0ex}}=3m\left(2m-3n\right)-4n\left(2m-3n\right)\phantom{\rule{0ex}{0ex}}=6{m}^{2}-9mn-8mn+12{n}^{2}\phantom{\rule{0ex}{0ex}}=6{m}^{2}-17mn+12{n}^{2}$

Page No 87:

By horizontal method:

$\left({x}^{2}-{a}^{2}\right)×\left(x-a\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(x-a\right)-{a}^{2}\left(x-a\right)\phantom{\rule{0ex}{0ex}}={x}^{3}-a{x}^{2}-{a}^{2}x+{a}^{3}$
i.e $\left({x}^{3}+{a}^{3}\right)-ax\left(x-a\right)$

Page No 87:

By horizontal method:

$\left({x}^{2}-{y}^{2}\right)×\left(x+2y\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(x+2y\right)-{y}^{2}\left(x+2y\right)\phantom{\rule{0ex}{0ex}}={x}^{3}+2{x}^{2}y-x{y}^{2}-2{y}^{3}\phantom{\rule{0ex}{0ex}}i.e\left({x}^{3}-2{y}^{3}\right)+xy\left(2x-y\right)$

Page No 87:

By horizontal method:

$\left(3{p}^{2}+{q}^{2}\right)×\left(2{p}^{2}-3{q}^{2}\right)\phantom{\rule{0ex}{0ex}}=3{p}^{2}\left(2{p}^{2}-3{q}^{2}\right)+{q}^{2}\left(2{p}^{2}-3{q}^{2}\right)\phantom{\rule{0ex}{0ex}}=6{p}^{4}-9{p}^{2}{q}^{2}+2{p}^{2}{q}^{2}-3{q}^{4}\phantom{\rule{0ex}{0ex}}i.e6{p}^{4}-7{p}^{2}{q}^{2}-3{q}^{4}$

Page No 87:

By horizontal method:

$\left(2{x}^{2}-5{y}^{2}\right)×\left({x}^{2}+3{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=2{x}^{2}\left({x}^{2}+3{y}^{2}\right)-5{y}^{2}\left({x}^{2}+3{y}^{2}\right)\phantom{\rule{0ex}{0ex}}=2{x}^{4}+6{x}^{2}{y}^{2}-5{x}^{2}{y}^{2}-15{y}^{4}\phantom{\rule{0ex}{0ex}}=2{x}^{4}+{x}^{2}{y}^{2}-15{y}^{4}\phantom{\rule{0ex}{0ex}}$

Page No 87:

By horizontal method:

$\left({x}^{3}-{y}^{3}\right)×\left({x}^{2}+{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{3}\left({x}^{2}+{y}^{2}\right)-{y}^{3}\left({x}^{2}+{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{5}+{x}^{3}{y}^{2}-{x}^{2}{y}^{3}-{y}^{5}\phantom{\rule{0ex}{0ex}}=\left({x}^{5}-{y}^{5}\right)+{x}^{2}{y}^{2}\left(x-y\right)$

Page No 87:

By horizontal method:
$\left({x}^{4}+{y}^{4}\right)×\left({x}^{2}-{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{4}\left({x}^{2}-{y}^{2}\right)+{y}^{4}\left({x}^{2}-{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{6}-{x}^{4}{y}^{2}+{y}^{4}{x}^{2}-{y}^{6}\phantom{\rule{0ex}{0ex}}=\left({x}^{6}-{y}^{6}\right)-{x}^{2}{y}^{2}\left({x}^{2}-{y}^{2}\right)$

Page No 87:

By horizontal method:

Page No 87:

By horizontal method:

$\left({x}^{2}-3x+7\right)×\left(2x+3\right)\phantom{\rule{0ex}{0ex}}=2x\left({x}^{2}-3x+7\right)+3\left({x}^{2}-3x+7\right)\phantom{\rule{0ex}{0ex}}=2{x}^{3}-6{x}^{2}+14x+3{x}^{2}-9x+21\phantom{\rule{0ex}{0ex}}=2{x}^{3}-3{x}^{2}+5x+21\phantom{\rule{0ex}{0ex}}$

Page No 87:

By horizontal method:
$\left(3{x}^{2}+5x-9\right)×\left(3x-5\right)\phantom{\rule{0ex}{0ex}}=3x\left(3{x}^{2}+5x-9\right)-5\left(3{x}^{2}+5x-9\right)\phantom{\rule{0ex}{0ex}}=9{x}^{3}+15{x}^{2}-27x-15{x}^{2}-25x+45\phantom{\rule{0ex}{0ex}}=9{x}^{3}-52x+45$

Page No 87:

By horizontal method:
$\left({x}^{2}-xy+{y}^{2}\right)×\left(x+y\right)\phantom{\rule{0ex}{0ex}}=x\left({x}^{2}-xy+{y}^{2}\right)+y\left({x}^{2}-xy+{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{3}-{x}^{2}y+{y}^{2}x+{x}^{2}y-x{y}^{2}+{y}^{3}\phantom{\rule{0ex}{0ex}}={x}^{3}+{y}^{3}$

Page No 87:

By horizontal method:

$\left({x}^{2}+xy+{y}^{2}\right)×\left(x-y\right)\phantom{\rule{0ex}{0ex}}x\left({x}^{2}+xy+{y}^{2}\right)-y\left({x}^{2}+xy+{y}^{2}\right)\phantom{\rule{0ex}{0ex}}={x}^{3}+{x}^{2}y+x{y}^{2}-{x}^{2}y-x{y}^{2}-{y}^{3}\phantom{\rule{0ex}{0ex}}={x}^{3}-{y}^{3}$

Page No 87:

By horizontal method:

$\left({x}^{3}-2{x}^{2}+5\right)×\left(4x-1\right)\phantom{\rule{0ex}{0ex}}=4x\left({x}^{3}-2{x}^{2}+5\right)-1\left({x}^{3}-2{x}^{2}+5\right)\phantom{\rule{0ex}{0ex}}=4{x}^{4}-8{x}^{3}+20x-{x}^{3}+2{x}^{2}-5\phantom{\rule{0ex}{0ex}}=4{x}^{4}-9{x}^{3}+2{x}^{2}+20x-5$

Page No 87:

By horizontal method:

$\left(9{x}^{2}-x+15\right)×\left({x}^{2}-3\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(9{x}^{2}-x+15\right)-3\left(9{x}^{2}-x+15\right)\phantom{\rule{0ex}{0ex}}=9{x}^{4}-{x}^{3}+15{x}^{2}-27{x}^{2}+3x-45\phantom{\rule{0ex}{0ex}}=9{x}^{4}-{x}^{3}-12{x}^{2}+3x-45$

Page No 87:

By horizontal method:

$\left({x}^{2}-5x+8\right)×\left({x}^{2}+2\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left({x}^{2}-5x+8\right)+2\left({x}^{2}-5x+8\right)\phantom{\rule{0ex}{0ex}}={x}^{4}-5{x}^{3}+8{x}^{2}+2{x}^{2}-10x+16\phantom{\rule{0ex}{0ex}}={x}^{4}-5{x}^{3}+10{x}^{2}-10x+16$

Page No 87:

By horizontal method:

$\left({x}^{3}-5{x}^{2}+3x+1\right)×\left({x}^{2}-3\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left({x}^{3}-5{x}^{2}+3x+1\right)-3\left({x}^{3}-5{x}^{2}+3x+1\right)\phantom{\rule{0ex}{0ex}}={x}^{5}-5{x}^{4}+3{x}^{3}+{x}^{2}-3{x}^{3}+15{x}^{2}-9x-3\phantom{\rule{0ex}{0ex}}={x}^{5}-5{x}^{4}+16{x}^{2}-9x-3$

Page No 87:

By horizontal method:

$\left(3x+2y-4\right)×\left(x-y+2\right)\phantom{\rule{0ex}{0ex}}x\left(3x+2y-4\right)-y\left(3x+2y-4\right)+2\left(3x+2y-4\right)\phantom{\rule{0ex}{0ex}}=3{x}^{2}+2xy-4x-3xy-2{y}^{2}+4y+6x+4y-8\phantom{\rule{0ex}{0ex}}=3{x}^{2}-2{y}^{2}-xy+2x+8y-8$

Page No 87:

By horizontal method:

$\left({x}^{2}-5x+8\right)×\left({x}^{2}+2x-3\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left({x}^{2}-5x+8\right)+2x\left({x}^{2}-5x+8\right)-3\left({x}^{2}-5x+8\right)\phantom{\rule{0ex}{0ex}}={x}^{4}-5{x}^{3}+8{x}^{2}+2{x}^{3}-10{x}^{2}+16x-3{x}^{2}+15x-24\phantom{\rule{0ex}{0ex}}={x}^{4}-3{x}^{3}-5{x}^{2}+31x-24$

Page No 87:

By horizontal method:

$\left(2{x}^{2}+3x-7\right)×\left(3{x}^{2}-5x+4\right)\phantom{\rule{0ex}{0ex}}=2{x}^{2}\left(3{x}^{2}-5x+4\right)+3x\left(3{x}^{2}-5x+4\right)-7\left(3{x}^{2}-5x+4\right)\phantom{\rule{0ex}{0ex}}=6{x}^{4}-10{x}^{3}+8{x}^{2}+9{x}^{3}-15{x}^{2}+12x-21{x}^{2}+35x-28\phantom{\rule{0ex}{0ex}}=6{x}^{4}-{x}^{3}-28{x}^{2}+47x-28$

Page No 87:

By horizontal method:

$\left(9{x}^{2}-x+15\right)×\left({x}^{2}-x-1\right)\phantom{\rule{0ex}{0ex}}={x}^{2}\left(9{x}^{2}-x+15\right)-x\left(9{x}^{2}-x+15\right)-1\left(9{x}^{2}-x+15\right)\phantom{\rule{0ex}{0ex}}=9{x}^{4}-{x}^{3}+15{x}^{2}-9{x}^{3}+{x}^{2}-15x-9{x}^{2}+x-15\phantom{\rule{0ex}{0ex}}=9{x}^{4}-10{x}^{3}+7{x}^{2}-14x-15$

Page No 90:

(i) 24x2y3 by 3xy

Therefore, the quotient is 8xy2.

(ii) 36xyz2 by −9xz

Therefore, the quotient is 4yz.

(iii)

Therefore, the quotient is 6xy.

(iv) −56mnp2 by 7mnp

Therefore, the quotient is −8p.

Page No 90:

(i) 5m3 − 30m2 + 45m by 5m

Therefore, the quotient is m2 6m + 9.

(ii) 8x2y2 − 6xy2 + 10x2y3 by 2xy

Therefore, the quotient is 4xy 3y + 5xy2.

(iii) 9x2y − 6xy + 12xy2 by − 3xy

Therefore, the quotient is −3x + 2 4y.

(iv) 12x4 + 8x3 − 6x2 by − 2x2

2-4x+32

Therefore the quotient is −6x2 4x + 3.

Page No 90:

Therefore, the quotient is $\left(x-2\right)$ and the remainder is 0.

Page No 90:

Therefore, the quotient is $x$−2 and the remainder is 0.

Page No 90:

(x2 + 12x + 35) by (x + 7)

Therefore, the quotient is $\left(x+5\right)$ and the remainder is 0.

Page No 90:

Therefore, the quotient is $\left(5x-3\right)$ and the remainder is 0.

Page No 90:

Therefore, the quotient is and the remainder is 0.

Page No 90:

Therefore, the quotient is and the remainder is 7.

Page No 90:

Therefore, the quotient is $\left({x}^{2}-x-1\right)$ and the remainder is 1.

Page No 90:

Therefore, the quotient is ${x}^{2}$-x+1 and the remainder is 0.

Page No 90:

Therefore, the quotient is ( x2 - 3x + 4) and remainder is 0.

Page No 90:

Therefore, the quotient is (x-1) and the remainder is 0.

Page No 90:

Therefore, the quotient is ( 5x+ 3) and the remainder is (x + 1).

Page No 90:

Therefore, the quotient is (x-1) and the remainder is 0.

Page No 90:

Therefore, the quotient is ( 4x2+ 3x -2) and the remainder is ( x-1).

(i) We have:

(ii) We have:

(iii) We have:

(iv) We have:

(v) We have:

(vi) We have:

(i) We have:

(ii) We have:

(iii) We have:

(iv) We have:

(v) We have:

(vi) We have:

Page No 93:

We shall use the identities (a+b)2 =a2 +b2 +2ab and (a-b)2 =a2 +b2 -2ab.

(i) We have:
$\left(8a+3b{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(8a\right)}^{2}+2×8a×3b+{\left(3b\right)}^{2}\phantom{\rule{0ex}{0ex}}=64{a}^{2}+48ab+9{b}^{2}$

(ii)We have:
$\left(7x+2y{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(7x\right)}^{2}+2×7x×2y+{\left(2y\right)}^{2}\phantom{\rule{0ex}{0ex}}=49{x}^{2}+28xy+4{y}^{2}$

(iii) We have :
$\left(5x+11{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(5x\right)}^{2}+2×5x×11+{\left(11\right)}^{2}\phantom{\rule{0ex}{0ex}}=25{x}^{2}+110x+121$

(iv) We have:
${\left(\frac{a}{2}+\frac{2}{a}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{a}{2}\right)}^{2}+2×\frac{a}{2}×\frac{2}{a}+{\left(\frac{2}{a}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\frac{a}{4}}^{2}+2+\frac{4}{{a}^{2}}$

(v) We have:
${\left(\frac{3x}{4}+\frac{2y}{9}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{3x}{4}\right)}^{2}+2×\frac{3x}{4}×\frac{2y}{9}+{\left(\frac{2y}{9}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\frac{9x}{16}}^{2}+\frac{1}{3}xy+\frac{4{y}^{2}}{81}$

(vi) We have:
$\left(9x-10{\right)}^{2}\phantom{\rule{0ex}{0ex}}{\left(9x\right)}^{2}-2×9x×10+{\left(10\right)}^{2}\phantom{\rule{0ex}{0ex}}=81{x}^{2}-180x+100$

(vii) We have:
$\left({x}^{2}y-y{z}^{2}{\right)}^{2}\phantom{\rule{0ex}{0ex}}{\left({x}^{2}y\right)}^{2}-2×{x}^{2}y×y{z}^{2}+{\left(y{z}^{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}={x}^{4}{y}^{2}-2{x}^{2}{y}^{2}{z}^{2}+{y}^{2}{z}^{4}$

(viii) We have:
${\left(\frac{x}{y}-\frac{y}{x}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{x}{y}\right)}^{2}-2×\frac{x}{y}×\frac{y}{x}+{\left(\frac{y}{x}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{{y}^{2}}-2+\frac{{y}^{2}}{{x}^{2}}$

(ix) We have:
${\left(3m-\frac{4}{5}n\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(3m\right)}^{2}-2×3m×\frac{4}{5}n+{\left(\frac{4}{5}n\right)}^{2}\phantom{\rule{0ex}{0ex}}=9{m}^{2}-\frac{24mn}{5}+\frac{16}{25}{n}^{2}$

(i) We have:

(ii) We have:

(iii) We have:

(iv) We have:

(v) We have:

(vi) We have:

(vii) We have:

(viii) We have:

(ix) We have:

Page No 94:

We shall use the identity (a+b)2 =a2 +b2 +2ab.

(i)
${\left(54\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(50+4{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(50\right)}^{2}+2×50×4+{\left(4\right)}^{2}\phantom{\rule{0ex}{0ex}}=2500+400+16\phantom{\rule{0ex}{0ex}}=2916$

(ii)
${\left(82\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(80+2{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(80\right)}^{2}+2×80×2+{\left(2\right)}^{2}\phantom{\rule{0ex}{0ex}}=6400+320+4\phantom{\rule{0ex}{0ex}}=6724$

(iii)
${\left(103\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(100+3{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(100\right)}^{2}+2×100×3+{\left(3\right)}^{2}\phantom{\rule{0ex}{0ex}}=10000+600+9\phantom{\rule{0ex}{0ex}}=10609$

(iv)
${\left(704\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(700+4{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(700\right)}^{2}+2×700×4+{\left(4\right)}^{2}\phantom{\rule{0ex}{0ex}}=490000+5600+16\phantom{\rule{0ex}{0ex}}=495616$

Page No 94:

We shall use the identity (a-b)2 = a2 +b2 -2ab.

(i)
${\left(69\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(70-1{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(70\right)}^{2}-2×70×1+1\phantom{\rule{0ex}{0ex}}=4900-140+1\phantom{\rule{0ex}{0ex}}=4761\phantom{\rule{0ex}{0ex}}$

(ii)
${\left(78\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(80-2{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(80\right)}^{2}-2×80×2+4\phantom{\rule{0ex}{0ex}}=6400-320+4\phantom{\rule{0ex}{0ex}}=6084\phantom{\rule{0ex}{0ex}}$

(iii)
${\left(197\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(200-3{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(200\right)}^{2}-2×200×3+9\phantom{\rule{0ex}{0ex}}=40000-1200+9\phantom{\rule{0ex}{0ex}}=38809\phantom{\rule{0ex}{0ex}}$

(iv)
${\left(999\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(1000-1{\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(1000\right)}^{2}-2×1000×1+1\phantom{\rule{0ex}{0ex}}=1000000-2000+1\phantom{\rule{0ex}{0ex}}=998001\phantom{\rule{0ex}{0ex}}$

Page No 94:

We shall use the identity (a-b) (a+b)=a2 - b2.

(i)
$\left(82{\right)}^{2}-\left(18{\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(82-18\right)\left(82+18\right)\phantom{\rule{0ex}{0ex}}=\left(64\right)\left(100\right)\phantom{\rule{0ex}{0ex}}=6400$

(ii)
$\left(128{\right)}^{2}-\left(72{\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(128-72\right)\left(128+72\right)\phantom{\rule{0ex}{0ex}}=\left(56\right)\left(200\right)\phantom{\rule{0ex}{0ex}}=11200$

(iii)
$197×203\phantom{\rule{0ex}{0ex}}=\left(200-3\right)\left(200+3\right)\phantom{\rule{0ex}{0ex}}={\left(200\right)}^{2}-{\left(3\right)}^{2}\phantom{\rule{0ex}{0ex}}=40000-9\phantom{\rule{0ex}{0ex}}=39991$

(iv)
$\frac{198×198-102×102}{96}\phantom{\rule{0ex}{0ex}}=\frac{{\left(198\right)}^{2}-{\left(102\right)}^{2}}{96}\phantom{\rule{0ex}{0ex}}=\frac{\left(198-102\right)\left(198+102\right)}{96}\phantom{\rule{0ex}{0ex}}=\frac{\left(96\right)\left(300\right)}{96}\phantom{\rule{0ex}{0ex}}=300$

(v)
$\left(14.7×15.3\right)\phantom{\rule{0ex}{0ex}}=\left(15-0.3\right)×\left(15+0.3\right)\phantom{\rule{0ex}{0ex}}=\left(15{\right)}^{2}-\left(0.3{\right)}^{2}\phantom{\rule{0ex}{0ex}}=225-0.09\phantom{\rule{0ex}{0ex}}=224.91$

(vi)
$\left(8.63{\right)}^{2}-\left(1.37{\right)}^{2}\phantom{\rule{0ex}{0ex}}=\left(8.63-1.37\right)\left(8.63+1.37\right)\phantom{\rule{0ex}{0ex}}=\left(7.26\right)\left(10\right)\phantom{\rule{0ex}{0ex}}=72.6$

Page No 94:

Therefore, the value of the expression (9x2 + 24x + 16), when x = 12, is 1600.

Page No 94:

Therefore, the value of the expression (64x2 + 81y2 + 144xy), when x = 11 and y = $\frac{4}{3}$, is 10000.y=43

Page No 94:

Therefore, the value of  x2+$\frac{1}{{x}^{2}}$ is 14.

Therefore, the value of x4 + $\frac{1}{{x}^{4}}$ is 194.

(c) (−6a + 17b)

Page No 95:

(d) (3p2 + 5q − 9r3 +7)

(d) x2 + 2x − 15

Page No 95:

(b) (6x2 + 7x − 3)

Page No 95:

(c) (x2 + 8x + 16)

Page No 95:

(d) (x2 − 12x + 36)

(b) (4x2 − 25)

(c) −4ab2

(b) (2x + 1)

(a) (x − 2)

(c) (a4 − 1)

Page No 95:

a) $\left(\frac{1}{{x}^{2}}-\frac{1}{{y}^{2}}\right)$

(1x21y2)

(c) 23

(b) 38

Page No 95:

(c) 6400

[using the identity (a-b)(a+b)=a2 -b2]

Page No 95:

(a) 39991

$\left(197\right)×\left(203\right)\phantom{\rule{0ex}{0ex}}⇒\left(200-3\right)\left(200+3\right)\phantom{\rule{0ex}{0ex}}⇒{\left(200\right)}^{2}-{\left(3\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒40000-9\phantom{\rule{0ex}{0ex}}⇒39991$                   [using the identity (a+b) (a-b) = a2  -b2]

(b) 116