Question

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(i) ${(2x-\frac{1}{3x})}^2$ (ii) ${(3x^2+5y)}^2$
(iii) $(\sqrt{2x}-3y)(\sqrt{2x}+3y)$ (iv) ${(\frac{1}{4}a-\frac{1}{2}b+1)}^2$

Answer 1

(i) ${(2x-\frac{1}{3x})}^2$
Using, $(a-b{)}^2=a^2-2ab+b^2$
$=(2x{)}^2-2(2x)\left(\frac{1}{3x}\right)+\frac{1}{(3x{)}^2}=4x^2-\frac{4}{3}+\frac{1}{9x^2}$
(ii) ${(3x^2+5y)}^2$
Using, $(a+b{)}^2=a^2+2ab+b^2$
$=(3x^2{)}^2+2(3x^2\left)\right(5y)+(5y{)}^2$
$=9x^4+30x^{2}y+25y^2$
(iii) $(\sqrt{2x}-3y)(\sqrt{2x}+3y)$
Using, $(a-b)(a+b)=a^2-b^2$
$=(\sqrt{2}x{)}^2-(3y{)}^2$
$=2x^2-9y^2$
(iv) ${(\frac{1}{4}a-\frac{1}{2}b+1)}^2$
Using, $(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$
$={\left(\frac{1}{4}a\right)}^2+{(-\frac{1}{2}b)}^2+(1{)}^2+2\left(\frac{1}{4}a\right)(-\frac{1}{2}b)+2(-\frac{1}{2}b)(1)+2(1)\left(\frac{1}{4}a\right)$
$=\frac{1}{16}{a}^2+\frac{1}{4}{b}^2+1-\frac{ab}{4}-b+\frac{a}{2}$