\(đkxđ:x\ne\pm1\)
\(\dfrac{x^2}{x-1}-\dfrac{6}{x^2-1}=\dfrac{6}{x+1}\\ \Leftrightarrow\dfrac{x^2\left(x+1\right)-6}{x^2-1}=\dfrac{6\left(x-1\right)}{x^2-1}\\ \Leftrightarrow x^2\left(x+1\right)-6=6\left(x-1\right)\\ \Leftrightarrow x^3+x^2-6-6x+6=0\\ \Leftrightarrow x^3+x^2-6x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\\x=0\end{matrix}\right.\left(t/m\right)\)
\(\dfrac{x^2}{x-1}-\dfrac{6}{x^2-1}=\dfrac{6}{x+1}\left(x\ne1;x\ne-1\right)\\ < =>\dfrac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{6}{\left(x-1\right)\left(x+1\right)}=\dfrac{6\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
suy ra
`x^3 +x^2-6=6x-6`
`<=>x^3 +x^2 -6x-6=6=0`
`<=>x^3 +x^2 -6x=0`
`<=>x(x^2 +x-6)=0`
`<=>x(x^2 +3x-2x-6)=0`
`<=>x[x(x+3)-2(x+3)]=0`
`<=>x(x+3)(x-2)=0`
\(< =>\left[{}\begin{matrix}x=0\\x+3=0\\x-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\left(tm\right)\\x=-3\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
