Đáp án:
$*》ĐKXĐ:\\x≥0\\2\sqrt{x}-2\neq0\\2\sqrt{x}+2\neq0\\1-x\neq0\\=>x≥0 \text{và} x\neq1\\*》\text{Rút gọn} B:\\B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\\B=\dfrac{1}{2(\sqrt{x}-1)}-\frac{1}{2(\sqrt{x}+1)}+\dfrac{-\sqrt{x}}{(\sqrt{x}+1)(\sqrt{x}-1)}\\B=\dfrac{(\sqrt{x}+1)-(\sqrt{x}-1)+(-\sqrt{x}).2}{2(\sqrt{x}+1)(\sqrt{x}-1)}=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2(\sqrt{x}+1)(\sqrt{x}-1)}\\B=\dfrac{2-2\sqrt{x}}{2(\sqrt{x}+1)(\sqrt{x}-1)}=\dfrac{-2(\sqrt{x}-1)}{2(\sqrt{x}+1)(\sqrt{x}-1)}\\B=\dfrac{-1}{\sqrt{x}+1}\\*》\text{Tìm giá trị của} x \text{để}\\|B|=\dfrac{1}{2}\\\text{-Vì theo ĐKXĐ nên ta có:}\\|B|=\dfrac{1}{2}<=>|\frac{-1}{\sqrt{x}+1}|=\dfrac{1}{2}\\<=>-(\dfrac{-1}{\sqrt{x}+1})=\dfrac{1}{2}\\=>x=1\\\text{Vậy} x=1 \text{là giá trị cần tìm}.\\\text{Chúc bạn học giỏi.xin câu trả lời hay nhất}.$
