ĐKXĐ: \(x\in R\)
\(pt\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x+2\right)\sqrt{x^2-x+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2+x+1-\left(x+2\right)\sqrt{x^2-x+1}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+x+1-\left(x+2\right)\sqrt{x^2-x+1}=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2-x+1-\left(x+2\right)\sqrt{x^2-x+1}+2x=0\)
\(\Leftrightarrow\left(\sqrt{x^2-x+1}-2\right)\left(\sqrt{x^2-x+1}-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+1}=2\\\sqrt{x^2-x+1}=x\end{matrix}\right.\)
TH1: \(\sqrt{x^2-x+1}=2\Leftrightarrow x=\frac{1\pm\sqrt{13}}{2}\)
TH2: \(\sqrt{x^2-x+1}=x\Leftrightarrow\left\{{}\begin{matrix}-x+1=0\\x\ge0\end{matrix}\right.\Leftrightarrow x=1\)
Vậy phương trình đã cho có nghiệm \(x=\frac{1\pm\sqrt{13}}{2};x=1\)
