a) \(A=\dfrac{1}{3}\sqrt{27}-\sqrt{12}+\sqrt[]{7+4\sqrt{3}}\)
\(=\dfrac{1}{3}\sqrt{9.3}-\sqrt{4.3}+\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{3}-2\sqrt{3}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\sqrt{3}-2\sqrt{3}+2+\sqrt{3}=2\)
ĐK: \(x>0;x\ne1\)
\(B=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt[]{x}-1\right)}\right):\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
\(=\dfrac{x-1}{x}\)
b) \(B< A\Leftrightarrow\dfrac{x-1}{x}< 2\)
\(\Leftrightarrow x-1< 2x\)
\(\Leftrightarrow x>-1\)
Vậy, B<A<=> x>-1
10:
a: \(A=\dfrac{1}{3}\cdot3\sqrt{3}-2\sqrt{3}+\sqrt{3}+1=\sqrt{3}+\sqrt{3}-2\sqrt{3}+1=1\)
b: \(B=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(1+\sqrt{x}\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\)
