Question

these 4 sums pleaseusing identity..

Answer 1

the question is solved...
see the attachments

Answer 2

$\huge\underline\mathfrak{Solution}$



$\textbf{Step-by-step Solution}$


$\textbf{Answer.1}$

${ \: (2x + 7y)}^{2}$




$\textbf{Identity used}$

${ \: (x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$


$4 {x}^{2} + 49 {y}^{2} + 28xy$




$\textbf{Answer.2}$

${ (\frac{1}{2}x + \frac{2}{3} y )}^{2}$


$\textbf{Using,same Identity}$

${ \: ( \frac{1}{2} )}^{2} + { \: (\frac{2}{3}) }^{2} + 2 \times \frac{1}{2} \times \frac{2}{3}$

$\frac{1}{4} {x}^{2} + \frac{4}{9} {y}^{2} + 2 \times \frac{1}{2} \times \frac{2}{3}$

$\frac{1}{4} {x}^{2} + \frac{4}{9} {y}^{2} + \frac{2}{3}$




$\textbf{Answer.3}$

${(3x + \frac{1}{2x} )}^{2}$

$\textbf{Using,same Identity}$

$9 {x}^{2} + \frac{1}{16} {x}^{2} + 2 \times 3x \times \frac{1}{2x}$

$9 {x }^{2} + \frac{1}{16} {y}^{2} + 3$

$144 {x}^{2} + {y}^{2} + 48 = 0$