Đáp án:
\(\dfrac{{\sqrt x - 2}}{{1 + \sqrt x }}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{{1 + \sqrt x - \sqrt x }}{{1 + \sqrt x }}:\dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right) - \left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) + \sqrt x + 2}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{1}{{1 + \sqrt x }}.\dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}{{x - 9 - x + 4 + \sqrt x + 2}}\\
= \dfrac{1}{{1 + \sqrt x }}.\dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x - 2} \right)}}{{\sqrt x - 3}}\\
= \dfrac{{\sqrt x - 2}}{{1 + \sqrt x }}
\end{array}\)
