Question

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

Answer 1

điều kiện xác định : \(x>0;x\ne1\)

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

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

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

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