ĐK: \(x\ge5\)
Đặt \(\sqrt{x-5}=a\left(a\ge0\right)\Rightarrow a^2+1=x-4\)
\(PT\Leftrightarrow a^2+1+a=7\)
\(\Leftrightarrow a^2+a-6=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\left(tm\right)\\a=-3\left(l\right)\end{matrix}\right.\)
\(a=2\Rightarrow\sqrt{x-5}=2\Rightarrow x-5=4\Rightarrow x=9\)(tm)
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