\(\dfrac{1}{x-1}+\dfrac{3x^2}{1-x^3}=\dfrac{2x}{x^2+x+1}\)\(\left(ĐKXĐ:x\ne1\right)\)
\(\Leftrightarrow\dfrac{x^2+x+1-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Rightarrow x^2+x+1-3x^2=2x^2-2x\)
\(\Leftrightarrow4x^2-3x-1=0\)
\(\Leftrightarrow\left(4x^2-4x+x-1\right)=0\)
\(\Leftrightarrow4x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(ktmđk\right)\\x=-0,25\left(tmđk\right)\end{matrix}\right.\)
\(\dfrac{1}{x-1}+\dfrac{3x^2}{1-x^3}=\dfrac{2x}{x^2+x+1}\)(ĐKXĐ: \(x\ne1\))
\(\Leftrightarrow\dfrac{x^2+x+1-3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Rightarrow x^2+x+1-3x^2=2x^2-2x\)
\(\Leftrightarrow4x^2-3x-1=0\)
\(\Leftrightarrow\left(4x^2-4x+x-1\right)=0\)
\(\Leftrightarrow4x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(KTM\right)\\x=-\dfrac{1}{4}\left(TM\right)\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{4}\right\}\)
