ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}\left(a+b\right)\left(1-ab\right)=1\\a^2-b^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)\left(1-ab\right)=1\\\left(a+b\right)\left(a-b\right)=1\end{matrix}\right.\)
\(\Rightarrow1-ab=a-b\)
\(\Leftrightarrow ab+a-\left(b+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(b+1\right)=0\)
\(\Leftrightarrow a=1\Rightarrow\sqrt{x+2}=1\Rightarrow x=-1\)
