Question

HELLO BRAINLY USERS,
I am asking again this please give answer of all questions as proper answer in attachment​

Answer 1

Answer:

1.(5a+2)(5a+2)

7a+7a

14a

2.(12x+2)(12x+2)

14x+14x

28x

Answer 2

Explanation:

$\rm(a)(5a +2)(5a +2)$

$\green{ \sf Formula : - }$

$\rm(a + b)(a + b) = {a}^{2} + 2ab + {b}^{2}$

Here,

$\rm a = 5a \: \: and \: \: b = 2$

$\rm {(5a)}^{2} + 2(5a)(2) + {2}^{2}$

$\rm25a + 20a + 4$

$\rm (b.)( \frac{3}{2} x - \frac{2}{3} y)( \frac{3}{2} x - \frac{2}{3} y)$

$\green{ \sf Formula : - }$

$\rm(a - b)(a - b) = {a}^{2} - 2ab + {b}^{2}$

Here,

$\rm a = \frac{3}{2} x \: \: and \: \: b = - \frac{2}{3} y$

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

$\rm \frac{9}{4 } {x}^{2} + \frac{2 \times 3 \times 2 \times xy}{6} + \frac{4}{9} {y}^{2}$

$\rm \frac{9}{4 } {x}^{2} + \frac{ \cancel{12}xy}{ \cancel6} + \frac{4}{9} {y}^{2}$

$\rm \frac{9}{4 } {x}^{2} + 2xy + \frac{4}{9} {y}^{2}$

$\rm(c.)(12x - 2)(12x + 2)$

$\green{ \sf Formula : - }$

$\rm(a - b)(a + b) = {a}^{2} - {b}^{2}$

Here,

$\rm a = {12x} \: \: and \: \: b = 2$

$\rm {(12x)}^{2} - {2}^{2}$

$\rm 144x - 4$

$\rm4(36x + 1)$

$\rm (d.) \: (9a + 2b)(9a - 2b)$

$\green{ \sf Formula : - }$

$\rm(a + b)(a - b) = {a}^{2} - {b}^{2}$

Here,

$\rm a = 9a \: \: and \: \: b = 2b$

$\rm ({9a)^{2} - {(2b)} }^{2}$

$\rm 81 {a}^{2} - 4 {b}^{2}$

$\rm(e.) \: ( \frac{3}{5} a - 4)( \frac{3}{5} a - 4)$

$\green{ \sf Formula : - }$

$\rm(a - b)(a - b) = {a}^{2} - 2ab + {b}^{2}$

$\rm\: a = \frac{3}{5} a \: \: and \: \: b = - 4$

$\rm {(\frac{3}{5} a)}^{2} - 2(\frac{3}{5} a)(4) + {4}^{2}$

$\rm {\frac{9}{25} a}^{2} - \frac{2 \times 3 \times 4 \times a}{5} + 16$

$\rm {\frac{9}{25} a}^{2} - \frac{2 4 a}{5} + 16$

$\rm(f.) \: ( \frac{4p}{3} + \frac{q}{5} )( \frac{4p}{3} + \frac{q}{5} )$

$\green{ \sf Formula : - }$

$\rm(a + b)(a + b) = {a}^{2} + 2ab + {b}^{2}$

Here,

$\rm \: a = \frac{4p}{3} \: \: and \: \: b = \frac{q}{5}$

$\rm { \frac{16p}{9} }^{2} + {\frac{2 \times 4p \times q}{3 \times 5}}+ {\frac{q}{25} }^{2} </p><p>$

$\rm { \frac{16p}{9} }^{2} + {\frac{8pq}{15}}+ {\frac{q}{25} }^{2} </p><p>$

$\rm (g.) \: (a - 16)(a + 2)$

$\rm \: a(a - 16) + 2(a - 16)$

$\rm {a}^{2} - 16a + 2a - 32$

$\rm {a}^{2} - 14a - 32$

$\rm (h.)(11x - 7)(11x + 7)$

$\green{ \sf Formula : - }$

$\rm(a - b)(a + b) = {a}^{2} - {b}^{2}$

$\rm a = 11x \: \: and \: \: b = 7$

$\rm ({11x)}^{2} - {7}^{2}$

$\rm121 {x}^{2} - 49$

I hope it is helpful