\(x-2\sqrt{x}-3=x-2\sqrt{x}-2-1\)
\(=\left(x-1\right)-2\left(\sqrt{x}+1\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-2\left(\sqrt{x}+1\right)\)
\(=\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)\)
ĐKXĐ : x ≥ 0
\(x-2\sqrt{x}-3=x+\sqrt{x}-3\sqrt{x}-3\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)\)
Ta có: \(x-2\sqrt{x}-3\)
\(=x-3\sqrt{x}+\sqrt{x}-3\)
\(=\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}-3\right)\)
\(=\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)