ĐKXĐ: \(x\ne2;x\ne-3\)
Ta có: \(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)
\(\Leftrightarrow\frac{1}{x-2}-\frac{6}{x+3}=\frac{-5}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow\left(x+3\right)-6\left(x-2\right)=-5\)
\(\Leftrightarrow x+3-6x+12=-5\)
\(\Leftrightarrow-5x+15=-5\)
\(\Leftrightarrow5x=20\)
hay x=4
Vậy: x=4