a)
ĐK: \(x\ge0;x\ne1\)
Khi đó:
\(B=\left(\dfrac{2x+1}{\sqrt{x}^3-1}-\dfrac{x+\sqrt{x}+1}{\sqrt{x}^3-1}\right):\left(\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{x-2}{x+\sqrt{x}+1}\right)\\ =\dfrac{2x+1-x-\sqrt{x}-1}{\sqrt{x}^3-1}:\dfrac{x+\sqrt{x}+1-x+2}{x+\sqrt{x}+1}\\ =\dfrac{x-\sqrt{x}}{\sqrt{x}^3-1}:\dfrac{\sqrt{x}+3}{x+\sqrt{x}+1}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{\sqrt{x}+3}\\ =\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
b)
\(2B< 1\\ \Leftrightarrow B< \dfrac{1}{2}\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+3}< \dfrac{1}{2}\\ \Leftrightarrow\dfrac{1}{2}\sqrt{x}+\dfrac{3}{2}-\sqrt{x}< 0\\ \Leftrightarrow\dfrac{3}{2}-\dfrac{1}{2}\sqrt{x}< 0\\ \Leftrightarrow\dfrac{1}{2}\sqrt{x}< \dfrac{3}{2}\\ \Leftrightarrow\sqrt{x}< 3\\ \Leftrightarrow x< 9\)
Kết hợp điều kiện, kết luận: \(\left\{{}\begin{matrix}x\ne1\\0\le x< 9\end{matrix}\right.\) thì `2B<1`
$HaNa$
