ĐKXĐ: \(x\ge-1\)
\(5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=2\left(x^2+2\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+b^2=x^2+2\)
Pt trở thành:
\(5ab=2\left(a^2+b^2\right)\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x^2-x+1}=\sqrt{x+1}\\\sqrt{x^2-x+1}=2\sqrt{x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2-x+1\right)=x+1\\x^2-x+1=4\left(x+1\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
