Đặt \(t=\sqrt{x}\Rightarrow t^2=x\)
Ta có: \(t^2-2t-3=t^2+t-3t-3=t\left(t+1\right)-3\left(t+1\right)=\left(t+1\right)\left(t-3\right)\)
hay: \(\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)\)
\(=\sqrt{x}-2\sqrt{x}+1-4\)
\(=\left(\sqrt{x}-1\right)^2-2^2\)
\(=\left(\sqrt{x}-1+2\right)\left(\sqrt{x}-1-2\right)\)
\(=\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)\)
cai dau tien la \(\sqrt{x^2}\)
chứ ko phải \(\sqrt{x}\)