ĐKXĐ : \(\left\{{}\begin{matrix}x\ne1\\x\ge0\\x\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có : \(B=\left(\frac{1}{1-\sqrt{x}}+\frac{1}{1+\sqrt{x}}\right):\left(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}\right)+\frac{1}{2\sqrt{x}}\)
=> \(B=\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right):\left(\frac{1+\sqrt{x}-1+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right)+\frac{1}{2\sqrt{x}}\)
=> \(B=\left(\frac{2}{1-x}\right):\left(\frac{2\sqrt{x}}{1-x}\right)+\frac{1}{2\sqrt{x}}=\frac{2\left(1-x\right)}{2\sqrt{x}\left(1-x\right)}+\frac{1}{2\sqrt{x}}\)
=> \(B=\frac{1}{\sqrt{x}}+\frac{1}{2\sqrt{x}}=\frac{2}{2\sqrt{x}}+\frac{1}{2\sqrt{x}}=\frac{3}{2\sqrt{x}}\)
Vậy ....