Bài 3:
a) Ta có: \(A=\dfrac{1}{3}\cdot\sqrt{45}-\sqrt{20}+\sqrt{9+4\sqrt{5}}\)
\(=\dfrac{1}{3}\cdot3\sqrt{5}-2\sqrt{5}+\sqrt{5}+2\)
\(=2\)
Ta có: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{x-\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b) Để A=3B thì \(3\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=2\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2}{3}\)
\(\Leftrightarrow3\sqrt{x}-3-2\sqrt{x}=0\)
\(\Leftrightarrow x=9\)(thỏa ĐK)