Đáp án:

$x > {\left( {\dfrac{{19 + \sqrt {409} }}{2}} \right)^2}$

Giải thích các bước giải:

 ĐK: $x>0; x\ne 9$

Ta có:

$\begin{array}{l}
D = \left( {\dfrac{{\sqrt x }}{{3 + \sqrt x }} + \dfrac{{x - 9}}{{9 - x}}} \right):\left( {\dfrac{{3\sqrt x  + 1}}{{x - 3\sqrt x }} - \dfrac{1}{{\sqrt x }}} \right)\\
 = \left( {\dfrac{{\sqrt x }}{{3 + \sqrt x }} - 1} \right):\left( {\dfrac{{3\sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  - 3} \right)}} - \dfrac{1}{{\sqrt x }}} \right)\\
 = \left( {\dfrac{{\sqrt x  - \left( {3 + \sqrt x } \right)}}{{3 + \sqrt x }}} \right):\left( {\dfrac{{3\sqrt x  + 1 - \left( {\sqrt x  - 3} \right)}}{{\sqrt x \left( {\sqrt x  - 3} \right)}}} \right)\\
 = \left( {\dfrac{{ - 3}}{{3 + \sqrt x }}} \right):\left( {\dfrac{{2\sqrt x  + 4}}{{\sqrt x \left( {\sqrt x  - 3} \right)}}} \right)\\
 = \dfrac{{ - 3}}{{3 + \sqrt x }}.\dfrac{{\sqrt x \left( {\sqrt x  - 3} \right)}}{{2\left( {\sqrt x  + 2} \right)}}\\
 = \dfrac{{ - 3\sqrt x \left( {\sqrt x  - 3} \right)}}{{2\left( {\sqrt x  + 3} \right)\left( {\sqrt x  + 2} \right)}}
\end{array}$

Để $D<-1$

$\begin{array}{l}
 \Leftrightarrow \dfrac{{ - 3\sqrt x \left( {\sqrt x  - 3} \right)}}{{2\left( {\sqrt x  + 3} \right)\left( {\sqrt x  + 2} \right)}} <  - 1\\
 \Leftrightarrow  - 3\sqrt x \left( {\sqrt x  - 3} \right) <  - 2\left( {\sqrt x  + 3} \right)\left( {\sqrt x  + 2} \right)\left( {Do:2\left( {\sqrt x  + 3} \right)\left( {\sqrt x  + 2} \right) > 0} \right)\\
 \Leftrightarrow  - 3x + 9\sqrt x  <  - 2\left( {x + 5\sqrt x  + 6} \right)\\
 \Leftrightarrow  - 3x + 9\sqrt x  <  - 2x - 10\sqrt x  - 12\\
 \Leftrightarrow x - 19\sqrt x  - 12 > 0\\
 \Leftrightarrow {\left( {\sqrt x } \right)^2} - 2\sqrt x .\dfrac{{19}}{2} + \dfrac{{361}}{4} - \dfrac{{409}}{4} > 0\\
 \Leftrightarrow {\left( {\sqrt x  - \dfrac{{19}}{2}} \right)^2} > \dfrac{{409}}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  - \dfrac{{19}}{2} > \dfrac{{\sqrt {409} }}{2}\\
\sqrt x  - \dfrac{{19}}{2} <  - \dfrac{{\sqrt {409} }}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\sqrt x  > \dfrac{{19 + \sqrt {409} }}{2}\left( c \right)\\
\sqrt x  < \dfrac{{19 - \sqrt {409} }}{2}\left( l \right)
\end{array} \right.\\
 \Leftrightarrow x > {\left( {\dfrac{{19 + \sqrt {409} }}{2}} \right)^2}
\end{array}$

Vậy $x > {\left( {\dfrac{{19 + \sqrt {409} }}{2}} \right)^2}$ thỏa mãn đề.