Question

Answer 1

3: Ta có: \(P=\dfrac{2x-2\sqrt{x}}{x-1}\)

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

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

Answer 2

1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\) \(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}=-10\sqrt{3}\)

2) Để \(A=\sqrt{12-4x}\) có nghĩa thì \(12-4x\ge0\Leftrightarrow x\le\dfrac{12}{4}\Leftrightarrow x\le3\)

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