a, ĐKXĐ :\(\left\{{}\begin{matrix}x\ge0\\2\sqrt{x}-2\ne0\\2\sqrt{x}+2\ne0\\1-x\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\\x\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}\sqrt{x}\ne1\\\sqrt{x}\ne-1\\x\ge0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có : \(B=\frac{1}{2\sqrt{x}-2}-\frac{1}{2\sqrt{x}+2}+\frac{\sqrt{x}}{1-x}\)
=> \(B=\frac{1}{2\left(\sqrt{x}-1\right)}-\frac{1}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{x-1}\)
=> \(B=\frac{\sqrt{x}+1}{2\left(x-1\right)}-\frac{\sqrt{x}-1}{2\left(x-1\right)}-\frac{2\sqrt{x}}{2\left(x-1\right)}\)
=> \(B=\frac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{2\left(x-1\right)}\)
=> \(B=\frac{2-2\sqrt{x}}{2\left(x-1\right)}=\frac{2\left(1-\sqrt{x}\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(-\sqrt{x}-1\right)}\)
=> \(B=\frac{1}{-\sqrt{x}-1}=\frac{-1}{\sqrt{x}+1}\)
b, Thay x = 3 vào biểu thức B ta được :
\(B=\frac{-1}{\sqrt{3}+1}=\frac{1-\sqrt{3}}{2}\)
