ĐKXĐ: \(x\ge0\)
Đặt \(\sqrt{x}=t\ge0\Rightarrow x=t^2\)
Pt trở thành:
\(t^2+t-1=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{-1+\sqrt{5}}{2}\\t=\dfrac{-1-\sqrt{5}}{2}< 9\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}=\dfrac{-1+\sqrt{5}}{2}\)
\(\Rightarrow x=\dfrac{3-\sqrt{5}}{2}\)