Đáp án:
a. \(\dfrac{2}{{\sqrt x + 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 0;x \ne 9\\
A = \left[ {\dfrac{{\sqrt x + 3 - \sqrt x + 3}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}} \right].\dfrac{{\sqrt x - 3}}{3}\\
= \dfrac{6}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\sqrt x - 3}}{3}\\
= \dfrac{2}{{\sqrt x + 3}}\\
b.A > \dfrac{1}{3}\\
\to \dfrac{2}{{\sqrt x + 3}} > \dfrac{1}{3}\\
\to \dfrac{{6 - \sqrt x - 3}}{{3\left( {\sqrt x + 3} \right)}} > 0\\
\to \dfrac{{3 - \sqrt x }}{{3\left( {\sqrt x + 3} \right)}} > 0\\
Do:x \ge 0 \to 3\left( {\sqrt x + 3} \right) > 0\left( {ld} \right)\\
Bpt \to 3 - \sqrt x > 0\\
\to \sqrt x < 3\\
\to 0 \le x < 9;x \ne 1
\end{array}\)
