Question

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

rút gọn

 

 

Answer 1

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