Điều kiện: \(\left\{\begin{matrix}x\left(x-1\right)\ge0\\x\left(x+2\right)\ge0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x\le0\text{∨}x\ge1\\x\le-2\text{∨}x\ge0\\x\ge0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=0\\x\ge1\end{matrix}\right.\)
Với \(x=0\) thì \(\left(\text{*}\right)\Leftrightarrow0=0\Rightarrow x=0\)là 1 nghiệm của \(\left(\text{*}\right)\).
Với \(x\ge1\) thì \(\left(\text{*}\right)\Leftrightarrow\sqrt{x}\left(\sqrt{x-1}+\sqrt{x+2}\right)=2\sqrt{x^2}\Leftrightarrow\sqrt{x-1}+\sqrt{x+2}=2\sqrt{x}\)
\(\Leftrightarrow x-1+x+2+2\sqrt{\left(x-1\right)\left(x+2\right)}=4x\Leftrightarrow\sqrt{\left(x-1\right)\left(x+2\right)}=x-\frac{1}{2}\)\(\Leftrightarrow\left\{\begin{matrix}x\ge\frac{1}{2}\\x^2+x-2=x^2-x+\frac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x\ge\frac{1}{2}\\x=\frac{9}{8}\end{matrix}\right.\Leftrightarrow x=\frac{9}{8}\left(N\right)\)
Vậy phương trình có hai nghiệm là \(x=0\text{∨}x=\frac{9}{8}.\)