Question

Rút gọn

A=\(\left[\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x-2\sqrt{x}+1}\right]\frac{x\sqrt{x}-x-\sqrt{x}+1}{\sqrt{x}}\)

Điều kiện \(x\ge0;x\ne1\)

Answer 1

\(A=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right)\cdot\frac{x\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\sqrt{x}}\\ =\left(\frac{x-3\sqrt{x}+2-x-3\sqrt{x}-2}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right)\cdot\frac{\left(x-1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}\\ =\frac{-6\sqrt{x}}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\cdot\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}{\sqrt{x}}=-6\)

Làm hơi tắt, mong bạn thông cảm :<