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