ĐKXĐ: \(x\ne-1;x\ne2\)
\(\dfrac{x^2-x}{x^2-x+1}-\dfrac{x^2-x+2}{x^2-x-2}=1\) (1)
\(\Leftrightarrow\dfrac{x^2-x}{x^2-x+1}-\dfrac{x^2-x+2}{x^2-x-2}-1=0\)
\(\Leftrightarrow\dfrac{\left(x^2-x-2\right)\left(x^2-x\right)-\left(x^2-x+1\right)\left(x^2-x+2\right)-\left(x^2-x+1\right)\left(x^2-x-2\right)}{\left(x^2-x+1\right)\left(x^2-x-2\right)}=0\)
\(\Leftrightarrow\dfrac{2x^3-5x^2+4x-x^4}{\left(x^2-x+1\right)\left(x^2-x-2\right)}=0\)
\(\Leftrightarrow2x^3-5x^2+4x-x^4=0\)
\(\Leftrightarrow x\left(2x^2-5x+4-x^3\right)=0\)
\(\Leftrightarrow x\left(-x^3+2x^2-5x+4\right)=0\)
\(\Leftrightarrow x\left(-x^3+x^2+x^2-x-4x+4\right)=0\)
\(\Leftrightarrow x\left[-\left(x-1\right)\right]\left(x^2-x+4\right)=0\)
\(\Leftrightarrow-x\left(x-1\right)\left(x^2-x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\x-1=0\\x^2-x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\left(đk:x\ne-1;x\ne2\right)\\x\notin R\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{0;1\right\}\)
