ĐKXĐ: x≠-3; x≠3; x≠6
Ta có: \(\frac{x}{x+3}-\frac{x-2}{x-6}=\frac{x+2}{x^2-9}\)
\(\Leftrightarrow\frac{x\cdot\left(x-3\right)\cdot\left(x-6\right)}{\left(x+3\right)\left(x-3\right)\left(x-6\right)}-\frac{\left(x-2\right)\left(x+3\right)\left(x-3\right)}{\left(x-6\right)\left(x+3\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-6\right)}{\left(x-3\right)\left(x+3\right)\left(x-6\right)}=0\)
\(\Leftrightarrow x^3-6x^2-3x^2+18x-\left(x^3-2x^2-9x+18\right)-\left(x^2-4x-12\right)=0\)
\(\Leftrightarrow x^3-6x^2-3x^2+18x-x^3+2x^2+9x-18-x^2+4x+12=0\)
\(\Leftrightarrow-8x^2+31x-6=0\)
Δ=\(31^2-4\cdot\left(-8\right)\cdot\left(-6\right)=769\)
Vì Δ>0
nên \(\left\{{}\begin{matrix}x_1=\frac{-31-\sqrt{769}}{2\cdot\left(-8\right)}=\frac{31+\sqrt{769}}{16}\\x_2=\frac{-31+\sqrt{769}}{2\cdot\left(-8\right)}=\frac{31-\sqrt{769}}{16}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{\frac{31+\sqrt{769}}{16};\frac{31-\sqrt{769}}{16}\right\}\)
