ĐKXĐ: \(x\ge-3\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+2}=a>0\\\sqrt{x+3}=b\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x^2-x-5=2a^2-3b^2\\2x^2+x+1=2a^2-b^2\end{matrix}\right.\)
\(\Rightarrow\left(2a^2-3b^2\right)a+\left(2a^2-b^2\right)b\)
\(\Leftrightarrow2a^3+2a^2b-3ab^2-b^3=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a^2+4ab+b^2\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow x^2+x+2=x+3\Leftrightarrow x^2=1\)