Question

Rút gọn biểu thức

\(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x}}+1\right)\)   (x>0)

Answer 1

Với `x>0` có:

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

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

Answer 2

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

Answer 3

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