Chắc là `=0` nhỉ :(((
\(\dfrac{x^2-3x+1}{\left(x-1\right)\left(x-2\right)}-\dfrac{x}{x-2}-\dfrac{x+2}{1-x}=0\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne2\end{matrix}\right.\)
Ta có : \(\dfrac{x^2-3x+1}{\left(x-1\right)\left(x-2\right)}-\dfrac{x}{x-2}-\dfrac{x+2}{1-x}=0\)
\(\Leftrightarrow\dfrac{x^2-3x+1}{\left(x-1\right)\left(x-2\right)}-\dfrac{x}{x-2}+\dfrac{x+2}{x-1}=0\)
\(\Leftrightarrow\dfrac{x^2-3x+1}{\left(x-1\right)\left(x-2\right)}-\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x-2\right)}+\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x-1\right)\left(x+2\right)}=0\)
`=> x^2 -3x+1 -x^2 +x + x^2 -2x+2x-4=0`
`<=>x^2-2x-3=0`
`<=>x^2 +x-3x-3=0`
`<=>x(x+1)-3(x+1)=0`
`<=>(x+1)(x-3)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{-1;3\right\}\)
