ĐKXĐ: x∉{2;-2}
Ta có: \(\frac{1-6x}{x-2}-\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x\left(3x-2\right)+1}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow-6x^2-11x+2-\left(9x^2-14x-8\right)-\left(3x^2-2x+1\right)=0\)
\(\Leftrightarrow-6x^2-11x+2-9x^2+14x+8-3x^2+2x-1=0\)
\(\Leftrightarrow-18x^2+5x+9=0\)