Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4\\
x \ne 9
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
a,\\
P = \left( {\dfrac{{\sqrt x + 2}}{{x - 5\sqrt x + 6}} - \dfrac{{\sqrt x + 2}}{{2 - \sqrt x }} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}}} \right):\dfrac{{2 - \sqrt x }}{{\sqrt x + 1}}\\
= \left( {\dfrac{{\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}} + \dfrac{{\sqrt x + 2}}{{\sqrt x - 2}} - \dfrac{{\sqrt x + 2}}{{\sqrt x - 3}}} \right):\dfrac{{2 - \sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{\sqrt x + 2 + \left( {\sqrt x + 2} \right).\left( {\sqrt x - 3} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}:\dfrac{{2 - \sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x + 2} \right).\left[ {1 + \left( {\sqrt x - 3} \right) - \left( {\sqrt x - 2} \right)} \right]}}{{\left( {\sqrt x - 2} \right).\left( {\sqrt x - 3} \right)}}:\dfrac{{2 - \sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{{\left( {\sqrt x + 2} \right).0}}{{\left( {\sqrt x - 2} \right).\left( {\sqrt x - 3} \right)}}:\dfrac{{2 - \sqrt x }}{{\sqrt x + 1}}\\
= 0
\end{array}\)
