ĐKXĐ: 2x-1 \(\ge\)0 \(\Leftrightarrow\)x\(\ge\)\(\frac{1}{2}\) 

  \(\frac{1}{3-x}\)\(\ge\)0 \(\Leftrightarrow\)3-x\(\le\)0\(\Leftrightarrow\)-x\(\le\)-3 \(\Leftrightarrow\)x\(\ge\)3

đK: \(\hept{\begin{cases}x\ge0\\x\ne9\\x\ne4\end{cases}}\)

đặt biểu thức là A

Ta có: \(A=\left[1-\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\frac{\left(\sqrt{x}-3\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)\left(2-\sqrt{x}\right)+9-x}{\left(2-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right]\)

\(=\left[\frac{3}{\sqrt{x}+3}\right]:\left[\frac{4\sqrt{x}-x-4}{\left(2-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right]\)

\(=\left[\frac{3}{\sqrt{x}+3}\right].\left[\frac{\left(2-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{-\left(2-\sqrt{x}\right)^2}\right]=\frac{3}{\sqrt{x}-2}\)

-11/abc 

\(\left(\frac{2\sqrt{x}+x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\frac{x-1}{x+\sqrt{x}+1}\)

\(=\left(\frac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\frac{x-1}{x+\sqrt{x}+1}\)

\(=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\times\frac{\left(x+\sqrt{x}+1\right)}{x-1}=\frac{1}{x-1}\)