a. B = \(\frac{2x^2+2x}{x^2-1}+\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\left(x\ge0;x\ne1;x\ne-1\right)\)
= \(\frac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\sqrt{x}-1-\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
= \(\frac{2x}{x-1}+\frac{-2}{x-1}=\frac{2x-2}{x-1}=\frac{2\left(x-1\right)}{x-1}=2\)
Vậy B = 2 đkxđ \(x\ge0;x\ne1;x\ne-1\)
b. Để \(\sqrt{x}-B< 0\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp đkxđ => \(0\le x< 4;x\ne1\)
Vậy để \(\sqrt{x}-B< 0\)thì \(0\le x< 4;x\ne1\)
