`~rai~`
\(a)ĐKXĐ:\begin{cases}x\sqrt{x}-1\ne 0\\x+\sqrt{x}+1\ne 0\\x\ge 0\end{cases}\\\Leftrightarrow \begin{cases}x\ne 1\\x\ge 0\end{cases}\\b)\text{Với x}\ge 0;x\ne 1,\text{ta có:}\\B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\\\quad=\left[\dfrac{2x+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\dfrac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}\right]\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\dfrac{\sqrt{x}(1+\sqrt{x})}{1+\sqrt{x}}\right)\\\quad=\dfrac{2x+1-x+\sqrt{x}}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\dfrac{1+x\sqrt{x}-\sqrt{x}-x}{1+\sqrt{x}}\\\quad=\dfrac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}.\dfrac{(x\sqrt{x}-\sqrt{x})-(x-1)}{1+\sqrt{x}}\\\quad=\dfrac{1}{\sqrt{x}-1}.\dfrac{\sqrt{x}(x-1)-(x-1)}{1+\sqrt{x}}\\\quad=\dfrac{1}{\sqrt{x}-1}.\dfrac{(\sqrt{x}-1)(x-1)}{1+\sqrt{x}}\\\quad=\dfrac{x-1}{1+\sqrt{x}}\\\quad=\dfrac{(\sqrt{x}-1)(\sqrt{x}+1)}{1+\sqrt{x}}\\\quad=\sqrt{x}-1.\\\text{Vậy với x}\ge 0;x\ne 1\quad\text{thì B=}\sqrt{x}-1.\\c)x=\dfrac{2-\sqrt{x}}{2}\\\Leftrightarrow x=\dfrac{4-2\sqrt{3}}{4}\\\Leftrightarrow x=\dfrac{(\sqrt{3}-1)^2}{4}\\\text{Thay x=}\dfrac{(\sqrt{3}-1)^2}{4}(t/m)\quad\text{vào biểu thức B được:}\\B=\sqrt{\dfrac{(\sqrt{3}-1)^2}{4}}-1\\\quad=\dfrac{\sqrt{3}-1}{2}-1\\\quad=\dfrac{-3+\sqrt{3}}{2}.\\\text{Vậy giá trị biểu thức B tại x=}\dfrac{2-\sqrt{3}}{2}là\dfrac{-3+\sqrt{3}}{2}.\)
