\(\frac{1}{2-x}+1=\frac{1}{x+2}-\frac{6-x}{3x^2-12}\)ĐKXĐ : \(x\ne\pm2\)
\(\Leftrightarrow\frac{-3\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}+\frac{3\left(x-2\right)\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}=\frac{3\left(x-2\right)}{3\left(x-2\right)\left(x+2\right)}+\frac{x-6}{3\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{-3x-6+3\left(x^2-4\right)}{3\left(x-2\right)\left(x+2\right)}-\frac{3x-6+x-6}{3\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{-3x-6+3x^2-12-3x+6-x+6}{3\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{-7x-6+3x^2}{3\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow3x^2-7x-6=0\)
\(\Leftrightarrow3x^2-9x+2x-6=0\)
\(\Leftrightarrow3x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-2}{3}\end{cases}}\)( thỏa mãn )
Vậy....
