ĐKXĐ: \(-\sqrt{5}\le x\le\sqrt{5}\)
\(x^2+\sqrt{x+5}=5\Leftrightarrow x+1-\sqrt{x+5}-x^2-x+4=0\Leftrightarrow\dfrac{x^2+x-4}{x+1+\sqrt{x+5}}-\left(x^2+x-4\right)=0\Leftrightarrow\left(x^2+x-4\right)\left(\dfrac{1}{x+1+\sqrt{x+5}}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-4=0\left(1\right)\\\dfrac{1}{x+1+\sqrt{x+5}}-1=0\left(2\right)\end{matrix}\right.\)
Từ (1) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1+\sqrt{17}}{2}\left(tmđk\right)\\x=\dfrac{-1-\sqrt{17}}{2}\left(ktmđk\right)\end{matrix}\right.\)
Từ (2) \(\Rightarrow\) \(-x-\sqrt{x+5}=0\) (quy đồng)
\(\Leftrightarrow\sqrt{x+5}=-x\left(x\le0\right)\Leftrightarrow x+5=x^2\Leftrightarrow x^2-x-5=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{21}}{2}\left(ktm\right)\\x=\dfrac{1-\sqrt{21}}{2}\left(tm\right)\end{matrix}\right.\)
Vậy nghiệm của pt là \(x=\dfrac{-1+\sqrt{17}}{2};x=\dfrac{1-\sqrt{21}}{2}\)
