Question

pls fast
que 1 and. 2
pls don't scam
i will mark u as brainlist​

Answer 1

2

Ans i. (14p) = 14p × 14p= 615p

Ans ii.

Answer 2

Step-by-step explanation:

Solution (i):-

$\begin{aligned} \text {LHS } & \rm=(3 x+7)^{2}-84 x \\ \\ & \rm=\left(9 x^{2}+42 x+49\right)-84 x \\ \\ & \rm=9 x^{2}-42 x+49 \\ \\ & \rm=(3 x)^{2}-2(3 x)(7)+(7)^{2} \\ \\ & \rm=(3 x-7)^{2}-\text { RHS } \end{aligned}$

Solution (ii):-

$\begin{aligned} \text { LHS } & \rm=(9 p-5 q)^{2}+180 p q \\ \\ & \rm=(9 p)^{2}-2(9 p)(5 q)+(5 q)^{2}+180 p q \\ \\ & \rm=81 p^{2}-90 p q+25 q^{2}+180 p q \\ \\ &\rm=81 p^{2}+(-90+180) p q+25 q^{2} \\ \\ &\rm=81 p^{2}+90 p q+25 q^{2} \\ \\ \text { RHS } &\rm=(9 p+5 q)^{2} \\ \\ &\rm=(9 p)^{2}+2(9 p)(5 q)+(5 q)^{2} \\ \\ &\rm=81 p^{2}+90 p q+25 q^{2} \\ \\ \text { Since, LHS } &=\text { RHS } \\ \\ \rm \therefore(9 p&\rm-5 q)^{2}+180 p q=(9 p+5 q)^{2} \end{aligned}$

Solution (iii):-

$\begin{aligned} \text { L.H.S. } & \rm=\left(\frac{4}{3} m\right)^{2}+\left(\frac{3}{4} n\right)^{2}-2 \times \frac{4}{3} m \times \frac{3}{4} n+2 m n \\ \\ &\rm=\frac{16}{9} m^{2}+\frac{9}{16} n^{2}-2 \times \frac{4}{3} \times \frac{3}{4} m n+2 m n \\ \\ &\rm=\frac{16}{9} m^{2}+\frac{9}{16} n^{2}-2 m n+2 m n \\ \\ &\rm=\left(\frac{16}{9} m^{2}+\frac{9}{16} n^{2}\right)=\text { R.H.S. } \end{aligned}$