Question

Answer 1

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

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

Để \(A=\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\)

\(\Rightarrow3\sqrt{x}-3=\sqrt{x}\Rightarrow2\sqrt{x}=3\)

\(\Rightarrow\sqrt{x}=\dfrac{3}{2}\Rightarrow x=\dfrac{9}{4}\)