\(\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}\)=\(\frac{1^2+2.1.\sqrt{x}+\left(\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}=\frac{1+2\sqrt{x}+x-4\sqrt{x}}{1-\sqrt{x}}=\frac{1-2\sqrt{x}+x}{1-\sqrt{x}}=\frac{1^2-2.1.\sqrt{x}+\left(\sqrt{x}\right)^2}{1-\sqrt{x}}=\frac{\left(1-\sqrt{x}\right)^2}{1-\sqrt{x}}=1-\sqrt{x}\)
\(\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}=\frac{1+2\sqrt{x}+x-4\sqrt{x}}{1-\sqrt{x}}\) (đk x\(\ge\)0,x\(\ne\)1)
\(=\frac{1-2\sqrt{x}+x}{1-\sqrt{x}}=\frac{\left(1-\sqrt{x}\right)^2}{1-\sqrt{x}}\)
\(=1-\sqrt{x}\)
Vậy \(\frac{\left(1+\sqrt{x}\right)^2-4\sqrt{x}}{1-\sqrt{x}}=1-\sqrt{x}\) (\(x\ge0,x\ne1\))
