ĐKXĐ:$x≥0$
$2\sqrt{x}-2\neq0$ $x≥0$
$2\sqrt{x}+2\neq0$ $<=>x\neq1$
$1-x\neq0$
Rút gọn $B$:
$B=\dfrac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}$
$B=\dfrac{1}{2(\sqrt{x}-1)}-\dfrac{1}{2(\sqrt{x}+1)}+\dfrac{-\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}$
$B=\dfrac{(\sqrt{x}+1)-(\sqrt{x}-1)+(-\sqrt{x}).2}{2(\sqrt{x}+1)(\sqrt{x}-1)}=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2(\sqrt{x}+1)(\sqrt{x}-1)}$
$B=\dfrac{2-2\sqrt{x}}{2(\sqrt{x}+1)(\sqrt{x}-1)}=\dfrac{-2(\sqrt{x}-1)}{2(\sqrt{x}+1)(\sqrt{x}-1)}$
$B=\dfrac{-1}{\sqrt{x}+1}$
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