a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x-1\ne0\\2\sqrt{x}+1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ne1\\\sqrt{x}\ne-\frac{1}{2}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có : \(A=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right).\frac{x-\sqrt{x}}{2\sqrt{x}+1}\)
=> \(A=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
=> \(A=\frac{\sqrt{x}+1\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
=> \(A=\frac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)\(=\frac{\sqrt{x}}{\sqrt{x}+1}\)
b, - Thay x = 9 vào biểu thức trên ta được : \(\frac{\sqrt{9}}{\sqrt{9}+1}=\frac{3}{4}\)
