ĐKXĐ: ...
\(\Leftrightarrow3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}=2x^2-3x+10\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-2x+4}=a>0\\\sqrt{x+2}=b\ge0\end{matrix}\right.\) \(\Rightarrow2x^2-3x+10=2a^2+b^2\)
Pt trở thành:
\(3ab=2a^2+b^2\Leftrightarrow\left(a-b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-2x+4}=\sqrt{x+2}\\2\sqrt{x^2-2x+4}=\sqrt{x+2}\end{matrix}\right.\)
\(\Leftrightarrow...\)