\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}ĐKXĐ:x\ne1\)
\(\frac{1}{x-1}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{4}{x^2+x+1}\)
\(x^2+x+1+2x^2-5=4\left(x-1\right)\)
\(3x^2+x-4=4x-4\)
\(3x^2+x-4-4x+4=0\)
\(3x^2-3x=0\)
\(3x\left(x-1\right)=0\)
\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) Theo ĐKXĐ =>x=0
\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\frac{1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
1+2x2-5=4x-4
2x2-4x=-1+5-4
2x(x-2)=0
2x=0 hoặc x-2=0
x=0 hoặc x=2