Question

Помогите решить пожалуйста

Answer 1

ОДЗ: $\displaystyle \tt x\ne2, \: \: x\ne-2$

$\displaystyle \tt 1-\frac{1}{2-x}=\frac{6-x}{3x^2-12}-\frac{1}{x-2}\\\\ \displaystyle \tt 1-\frac{1}{2-x}-\frac{6-x}{3x^2-12}+\frac{1}{x-2}=0\\\\ \displaystyle \tt 1-\frac{1}{-(x-2)}-\frac{6-x}{3(x^2-4)}+\frac{1}{x-2}=0\\\\ \displaystyle \tt 1+\frac{1}{x-2}-\frac{6-x}{3(x-2)(x+2)}+\frac{1}{x-2}=0\\\\ \displaystyle \tt \frac{3(x-2)(x+2)+3(x+2)-(6-x)+3(x+2)}{3(x-2)(x+2)}=0\\\\ \displaystyle \tt \frac{3(x^2-4)+3x+6-6+x+3x+6}{3(x-2)(x+2)}=0$

$\displaystyle \tt \frac{3x^2-12+3x+6-6+x+3x+6}{3(x-2)(x+2)}=0\\\\ \displaystyle \tt \frac{3x^2-6+7x}{3(x-2)(x+2)}=0\\\\ \displaystyle \tt 3x^2-6+7x=0\\\\ \displaystyle \tt 3x^2+7x-6=0\\\\ \displaystyle \tt D=7^2-4\cdot3\cdot(-6)=49+72=121\\\\ \displaystyle \tt \sqrt{D}=\sqrt{121}=11\\\\ \displaystyle \tt \bold{x_1}=\frac{-7+11}{2\cdot3}=\frac{4}{6}=\frac{2}{3}\\\\ \displaystyle \tt \bold{x_2}=\frac{-7-11}{2\cdot3}=\frac{-18}{6}=-3$