ĐKXĐ: \(x\ge0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+7x+1}=b>0\\\sqrt{x}=c\ge0\end{matrix}\right.\) \(\Rightarrow b^2-a^2=8c^2\)
Ta được hệ: \(\left\{{}\begin{matrix}a+b=4c\\b^2-a^2=8c^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4c\\\left(a+b\right)\left(b-a\right)=8c^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=4c\\b-a=2c\end{matrix}\right.\) \(\Rightarrow b=3c\)
\(\Leftrightarrow\sqrt{x^2+7x+1}=3\sqrt{x}\)
\(\Leftrightarrow x^2+7x+1=9x\)
\(\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\)
