Question

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

Rút gọn

Answer 1

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

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

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

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

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

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

= -x + \(\sqrt{x}\)

Answer 2

\(x\ge0;x\ne\pm1\)

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

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

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

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