\(x-\sqrt{x}-2\)
\(=x-\sqrt{x}+\frac{1}{4}-\frac{9}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\left(\frac{3}{2}\right)^2\)
\(=\left(\sqrt{x}-\frac{1}{2}-\frac{3}{2}\right)\left(\sqrt{x}-\frac{1}{2}+\frac{3}{2}\right)\)
\(=\sqrt{x-2}\left(\sqrt{x}+1\right)\)
Không biết đúng không nữa
\(x-\sqrt{x}-2=0\) (ĐKXĐ : \(x\ge0\))
\(\Leftrightarrow\left(x+\sqrt{x}\right)-\left(2\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}-2=0\\\sqrt{x}+1=0\end{array}\right.\)\(\Leftrightarrow\sqrt{x}-2=0\) (vì \(\sqrt{x}+1>0\) )
\(\Leftrightarrow x=4\) (TMĐK)
Vậy phương trình có nghiệm x = 4
