Question

Помагите в фото Срочно !!!!!!!

Answer 1

$\displaystyle 1)(\frac{5}{x-2}-x-2)*\frac{2-x}{x^2-6x+9}=\frac{5-x(x-2)-2(x-2)}{x-2}*\frac{-(x-2)}{(x-3)^2}=\frac{5-x^2+2x-2x+4}{x-2}*\\ \\ \frac{-(x-2)}{(x-3)^2}=\frac{9-x^2}{x-2}*\frac{-(x-2)}{(x-3)^2}=\frac{(3-x)(3+x)}{x-2}*\frac{-(x-2)}{(x-3)^2}=-(x-3)(3+x)*\\ \\ \\ \frac{-1}{(x-3)^2}=-(3+x)*\frac{-1}{x-3}=\frac{3+x}{x-3}\\\\ 2)x+2=\frac{3}{x},x\neq 0\\(x+2)*x=3\\(x+2)*x-3=0\\x^2+2x-3=0\\x^2+3x-x-3=0\\x(x+3)-(x+3)=0 \\(x+3)(x-1)=0\\x+3=0,x-1=0\\x_1=-3,x_2=1$