a, D xác định

\(\Leftrightarrow\hept{\begin{cases}x\ge0;\sqrt{x}\ne0\\9-x\ne0\\x-3\sqrt{x}\ne0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>0\\x\ne9\\x\ne0,x\ne9\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x>0\\x\ne9\end{cases}}\)

Vậy ...

b, \(D=\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)

\(D=\left[\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)}\right]:\left[\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)

\(D=\left[\frac{3\sqrt{x}-x+x+9}{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)}\right]:\left[\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)

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

\(D=\frac{3}{3-\sqrt{x}}\cdot\left(-\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\right)\)

\(D=-\frac{3\sqrt{x}}{2\sqrt{x}+4}\)

Vậy ...

c, \(D< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}< -1\left(x>0;x\ne9\right)\)

\(\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+\frac{2\sqrt{x}+4}{2\sqrt{x}+4}< 0\)

\(\Leftrightarrow\frac{-\sqrt{x}+4}{2\sqrt{x}+4}< 0\)

\(\Leftrightarrow-\sqrt{x}+4< 0\)

\(\Leftrightarrow\sqrt{x}>4\)

\(\Leftrightarrow x>16\left(tm\right)\)

Vậy ...

hàm số đồng biến khi \(a>0\)

\(< =>m>0\)thì hàm số dồng biến

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