a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có: \(C=\left(\frac{2\sqrt{x}}{\sqrt{x}-1}-\frac{x+1}{x\sqrt{x}-1}:\frac{\sqrt{x}}{\sqrt{x}-1}\right)\cdot\frac{\sqrt{x}}{x\sqrt{x}-1}\)
\(=\left(\frac{2\sqrt{x}}{\sqrt{x}-1}-\frac{x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}-1}{\sqrt{x}}\right)\cdot\frac{\sqrt{x}}{x\sqrt{x}-1}\)
\(=\left(\frac{2\sqrt{x}}{\sqrt{x}-1}-\frac{x+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{\sqrt{x}}{x\sqrt{x}-1}\)
\(=\left(\frac{2x\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\frac{\left(x+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{\sqrt{x}}{x\sqrt{x}-1}\)
\(=\frac{2x^2+2x\sqrt{x}+2x-\left(x\sqrt{x}-x+\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}}{x\sqrt{x}-1}\)
\(=\frac{2x^2+2x\sqrt{x}+2x-x\sqrt{x}+x-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}}{x\sqrt{x}-1}\)
\(=\frac{2x^2+x\sqrt{x}+3x-\sqrt{x}+1}{x^3-2x\sqrt{x}+1}\)