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\(\Leftrightarrow5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=2\left(x^2+2\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow5ab=2\left(a^2+b^2\right)\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\b=2a\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^2=4b^2\\b^2=4a^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=4\left(x+1\right)\\x+1=4\left(x^2-x+1\right)\end{matrix}\right.\) (casio)