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