Đáp án:
\( - \dfrac{3}{{\sqrt x + 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne 9\\
\left[ {\dfrac{{2\sqrt x \left( {\sqrt x - 3} \right) + \sqrt x \left( {\sqrt x + 3} \right) - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right]:\dfrac{{2\sqrt x - 2 - \sqrt x + 3}}{{\sqrt x - 3}}\\
= \dfrac{{2x - 6\sqrt x + x + 3\sqrt x - 3x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= \dfrac{{ - 3\sqrt x - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}\\
= - \dfrac{3}{{\sqrt x + 3}}
\end{array}\)
