ĐKXĐ: \(x\ge-2\)
Do \(\sqrt{x+5}+\sqrt{x+2}>0;\forall x\ge-2\) nên pt tương đương:
\(\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(\sqrt{x+5}+\sqrt{x+2}\right)\left(1+\sqrt{x^2+7x+10}\right)=3\left(\sqrt{x+5}+\sqrt{x+2}\right)\)
\(\Leftrightarrow3\left(1+\sqrt{x^2+7x+10}\right)=3\left(\sqrt{x+5}+\sqrt{x+2}\right)\)
\(\Leftrightarrow1+\sqrt{\left(x+5\right)\left(x+2\right)}=\sqrt{x+5}+\sqrt{x+2}\)
\(\Leftrightarrow\sqrt{\left(x+5\right)\left(x+2\right)}-\sqrt{x+2}-\left(\sqrt{x+5}-1\right)=0\)
\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{x+5}-1\right)-\left(\sqrt{x+5}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+5}-1\right)\left(\sqrt{x+2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1\\\sqrt{x+2}=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=-1\end{matrix}\right.\)
