\(ĐK:x\ge-1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\\\sqrt{x^2-x+1}=b\end{matrix}\right.\left(a,b\ge0\right)\)
\(PT\Leftrightarrow b^2-1+2ab=2a\\ \Leftrightarrow2ab-2a+b^2-1=0\\ \Leftrightarrow2a\left(b-1\right)+\left(b-1\right)\left(b+1\right)=0\\ \Leftrightarrow\left(2a+b+1\right)\left(b-1\right)=0\\ \Leftrightarrow b-1=0\left(2a+b+1>0\right)\\ \Leftrightarrow b=1\\ \Leftrightarrow x^2-x+1=1\\ \Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)