ĐKXĐ: \(x\ge-7\)
Đặt \(\sqrt{x+7}=a\ge0\Rightarrow7=a^2-x\) pt trở thành:
\(x^2+a=a^2-x\Leftrightarrow x^2-a^2+a+x=0\)
\(\Leftrightarrow\left(x-a\right)\left(x+a\right)+x+a=0\)
\(\Leftrightarrow\left(x+a\right)\left(x-a+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-x\\a=x+1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+7}=-x\left(x\le0\right)\\\sqrt{x+7}=x+1\left(x\ge-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-7=0\\x^2+x-6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{1+\sqrt{29}}{2}>0\left(l\right)\\x=\frac{1-\sqrt{29}}{2}\\x=2\\x=-3< -1\left(l\right)\end{matrix}\right.\)
