Question

Второе задание. Алгебра.

Answer 1

$frac{x^2+3x}{(x-3)^2}:left(frac3{x+3}+frac{x^2+9}{x^2-9}-frac3{3-x}right)=\=frac{x^2+3x}{(x-3)^2}:left(frac3{x+3}+frac{x^2+9}{(x-3)(x+3)}+frac3{x-3}right)=\=frac{x^2+3x}{(x-3)^2}:left(frac{3(x-3)+x^2+9+3(x+3)}{(x-3)(x+3)}right)=frac{x^2+3x}{(x-3)^2}:left(frac{3x-9+x^2+9+3x+9}{(x-3)(x+3)}right)=\=frac{x(x+3)}{(x-3)^2}cdotfrac{(x-3)(x+3)}{x^2+6x+9}=frac{x(x+3)}{(x-3)^2}cdotfrac{(x-3)(x+3)}{(x+3)^2}=frac x{x-3}\x=-299\frac{-299}{-299-3}=frac{299}{302}$