a) Ta có: \( \displaystyle C = \left( {{{\sqrt x } \over {3 + \sqrt x }} + {{x + 9} \over {9 - x}}} \right)\)\( \displaystyle :\left( {{{3\sqrt x + 1} \over {x - 3\sqrt x }} - {1 \over {\sqrt x }}} \right) \) \( \displaystyle = \left[ {{{\sqrt x } \over {3 + \sqrt x }} + {{x + 9} \over {\left( {3 + \sqrt x } \right)\left( {3 - \sqrt x } \right)}}} \right]\)\( \displaystyle :\left[ {{{3\sqrt x + 1} \over {\sqrt x \left( {\sqrt x - 3} \right)}} - {1 \over {\sqrt x }}} \right] \)\( \displaystyle = {{\sqrt x \left( {3 - \sqrt x } \right) + x + 9} \over {\left( {3 + \sqrt x } \right)\left( {3 - \sqrt x } \right)}}\)\( \displaystyle :{{3\sqrt x + 1 - \left( {\sqrt x - 3} \right)} \over {\sqrt x \left( {\sqrt x - 3} \right)}} \)\( \displaystyle = {{3\sqrt x - x + x + 9} \over {\left( {3 + \sqrt x } \right)\left( {3 - \sqrt x } \right)}}\)\( \displaystyle :{{2\sqrt x + 4} \over {\sqrt x \left( {\sqrt x - 3} \right)}} \) \( \displaystyle = {{3\sqrt x + 9} \over {\left( {3 + \sqrt x } \right)\left( {3 - \sqrt x } \right)}}\)\( \displaystyle .{{\sqrt x \left( {\sqrt x - 3} \right)} \over {2\sqrt x + 4}} \) \( \displaystyle = {{3\left( {\sqrt x + 3} \right)} \over {\left( {3 + \sqrt x } \right)\left( {3 - \sqrt x } \right)}}.{{ - \sqrt x \left( {3 - \sqrt x } \right)} \over {2\sqrt x + 4}} \)\( \displaystyle = \dfrac{{ - 3\sqrt x }}{{2\sqrt x + 4}}\) (với \( \displaystyle x > 0\) và \( \displaystyle x \ne 9)\)
b) Với \( \displaystyle C < - 1\) ta có: \( \displaystyle \begin{array}{l}\dfrac{{ - 3\sqrt x }}{{2\sqrt x + 4}} < - 1\\ \Leftrightarrow \dfrac{{ - 3\sqrt x }}{{2\sqrt x + 4}} + 1 < 0\\\Leftrightarrow \dfrac{{ - 3\sqrt x + 2\sqrt x + 4}}{{2\sqrt x + 4}} < 0\\\Leftrightarrow \dfrac{{4 - \sqrt x }}{{2\sqrt x + 4}} < 0\end{array}\)Vì \( \displaystyle x > 0\) nên \( \displaystyle \sqrt x > 0\)Khi đó: \( \displaystyle 2\sqrt x + 4 > 0\)Suy ra: \( \displaystyle 4 - \sqrt x < 0 \Leftrightarrow \sqrt x > 4 \Leftrightarrow x > 16\)Vậy với \( \displaystyle x > 16\) thì \( \displaystyle C < -1\).
