Question

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

Answer 1

Ta có:

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

Answer 2

ĐKXĐ: \(x>0\)

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

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

Answer 3

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

Answer 4

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

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

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