Question

Giúp mình với

Answer 1

\(a,A=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ b,x=16\Leftrightarrow A=\dfrac{4-1}{4}=\dfrac{3}{4}\\ c,A=\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{1}{3}\\ \Leftrightarrow3\sqrt{x}-3=\sqrt{x}\\ \Leftrightarrow2\sqrt{x}=3\\ \Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\left(tm\right)\)