Question

Tính A=?

A = (\(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\))/\(\dfrac{x+\sqrt{x}+1}{x\sqrt{x}-1}\)

Answer 1

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