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