\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(\frac{x+2}{x+1}-\frac{1}{x-2}=1-\frac{3}{x^2-x-2}\)
\(\frac{1}{x}+2=\left(\frac{1}{x}+2\right)\left(x^2+1\right)\)
\(\left(x+1+\frac{1}{x}\right)^2=\left(x-1-\frac{1}{x}\right)^2\)
\(\frac{x+9}{10}+\frac{x+10}{9}=\frac{9}{x+10}+\frac{10}{x+9}\)
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\left(x\ne1\right)\)
\(\Leftrightarrow\frac{1}{x-1}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x}{x^2+x+1}=0\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\left(x^2+x+1-3x^2-2x^2+2x\right)=0\)
\(\Leftrightarrow-4x^2+3x+1=0\left(\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}\ne0\right)\)
\(\Leftrightarrow-4x^2+4x-x+1=0\)
\(\Leftrightarrow-4x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-4x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\-4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\-4x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\left(loại\right)\\x=\frac{-1}{4}\end{cases}}}\)
Vậy \(x=\frac{-1}{4}\)