Question

Rút gọn biểu thức:

\(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}\)

Answer 1

\(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}\)

\(=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(1-x\right)^2}{2}\)

\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(1-x\right)\left(-\sqrt{x}-1\right)}.\dfrac{\left(1-x\right)^2}{2}\)

\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{-\sqrt{x}-1}.\dfrac{1-x}{2}\)

\(=\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\sqrt{x}+1}=\sqrt{x}-x\)

Answer 2

Ta có: \(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)

\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)^2}{2}\)

\(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}-2}{1}\cdot\dfrac{\sqrt{x}-1}{2}\)

\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{2}\)

\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)