Đáp án:
c) x>16
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x > 0;x \ne 9\\
b)C = \left[ {\dfrac{{\sqrt x \left( {\sqrt x - 3} \right) - x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right]:\left[ {\dfrac{{3\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 3} \right)}} - \dfrac{1}{{\sqrt x }}} \right]\\
= \dfrac{{x - 3\sqrt x - x - 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\left[ {\dfrac{{3\sqrt x + 1 - \sqrt x + 3}}{{\sqrt x \left( {\sqrt x - 3} \right)}}} \right]\\
= \dfrac{{ - 3\left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{2\sqrt x + 4}}\\
= - \dfrac{{3\sqrt x }}{{2\sqrt x + 4}}\\
c)C < - 1\\
\to - \dfrac{{3\sqrt x }}{{2\sqrt x + 4}} < - 1\\
\to \dfrac{{3\sqrt x }}{{2\sqrt x + 4}} > 1\\
\to \dfrac{{3\sqrt x - 2\sqrt x - 4}}{{2\sqrt x + 4}} > 0\\
\to \sqrt x - 4 > 0\left( {do:2\sqrt x + 4 > 0\forall x > 0} \right)\\
\to x > 16
\end{array}\)
