- ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x^2+x+1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ne0\end{matrix}\right.\)
=> \(x\ne1\)
- Ta có : \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
=> \(\frac{x^2+x+1}{x^3-1}+\frac{2x^2-5}{x^3-1}=\frac{4\left(x-1\right)}{x^3-1}\)
=> \(x^2+x+1+2x^2-5=4\left(x-1\right)\)
=> \(x^2+x+1+2x^2-5-4x+4=0\)
=> \(3x^2-3x=0\)
=> \(3x\left(x-1\right)=0\)
=> \(\left[{}\begin{matrix}3x=0\\x-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\left(TM\right)\\x=1\left(KTM\right)\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{0\right\}\)
