a) \(B=\left(\dfrac{\sqrt{x}}{x-1}-\dfrac{\sqrt{x}-1}{x+2\sqrt{x}+1}\right)\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}}{x-1}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}}{x-1}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\cdot\dfrac{\left(\sqrt{x}+1\right)^2}{3\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}+1}{3\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)}\)
\(=\dfrac{3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(3\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-1}\)
b) \(B< 0\Leftrightarrow\dfrac{1}{\sqrt{x}-1}< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow0\le x< 1\)
Kl:.....
