\(\dfrac{x^2-1}{2x^2-3x+1}=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(2x^2-2x\right)-\left(x-1\right)}=\dfrac{\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)-\left(x-1\right)}=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(2x-1\right)\left(x-1\right)}=\dfrac{x+1}{2x-1}\)
\(\left|2x+1\right|=3\\ \Rightarrow\left[{}\begin{matrix}2x+1=-3\\2x+1=3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
với \(x=-2\Rightarrow\dfrac{x+1}{2x-1}=\dfrac{-2+1}{2.\left(-2\right)-1}=\dfrac{1}{5}\)
với \(x=1\Rightarrow\dfrac{x+1}{2x-1}=\dfrac{1+1}{2.1-1}=2\)
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