Đáp án:
b) \(\dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\
b)P = \dfrac{{2\sqrt x - 9 + \left( {2\sqrt x + 1} \right)\left( {\sqrt x - 2} \right) - \left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{2\sqrt x - 9 + 2x - 3\sqrt x - 2 - x + 9}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - \sqrt x - 2}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}} = \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}
\end{array}\)
