Question

Решите пожалуйста очень надо

Answer 1

$1)\;\frac{16-x^2}{x^2+10x+25}:\frac{ax-4a}{3x+15}=\frac{(4-x)(4+x)}{(x+5)^2}:\frac{-a(4-x)}{3(x+5)}=-\frac{(4-x)(4+x)}{(x+5)^2}\cdot\frac{3(x+5)}{a(4-x)}=-\frac{3(4+x)}{a(x+5)}\\\\\\2)\;\frac{25-x^2}{x^2+2x+1}:\frac{bx-5b}{4x+4}=\frac{(5-x)(5+x)}{(x+1)^2}:\frac{-b(5-x)}{4(x+1)}=-\frac{(5-x)(5+x)}{(x+1)^2}\cdot\frac{4(x+1)}{b(5-x)}=-\frac{4(5+x)}{b(x+1)}$