ĐKXĐ: x≠2; x≠-2
Ta có: \(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\frac{1-6x}{x-2}+\frac{9x+4}{x+2}-\frac{x\left(3x-2\right)+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow-6x^2-11x+2+9x^2-14x-8-3x^2+2x-1=0\)
\(\Leftrightarrow-23x-7=0\)
\(\Leftrightarrow-23x=7\)
hay \(x=\frac{-7}{23}\)(tm)
Vậy: \(x=\frac{-7}{23}\)
