ĐKXĐ: \(x\ne1\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x}{x^2+x+1}\)
\(\Leftrightarrow\dfrac{x^2+x+1-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Rightarrow-2x^2+x+1=2x^2-2x\)
\(\Leftrightarrow-4x^2+3x+1=0\)
\(\Leftrightarrow4x^2-3x-1=0\)
\(\Leftrightarrow4x^2-4x+x-1=0\)
\(\Leftrightarrow4x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{4}\end{matrix}\right.\)
Vì \(x\ne1\) \(\Rightarrow x=-\dfrac{1}{4}\)
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