Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
A = \dfrac{{\sqrt x + 4}}{{\sqrt x + 2}} = \dfrac{{\sqrt {36} + 4}}{{\sqrt {36} + 2}} = \dfrac{{6 + 4}}{{6 + 2}} = \dfrac{{10}}{8} = \dfrac{5}{4}\\
2,\\
B = \left( {\dfrac{{\sqrt x }}{{\sqrt x + 4}} + \dfrac{4}{{\sqrt x - 4}}} \right):\dfrac{{x + 16}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 4} \right) + 4.\left( {\sqrt x + 4} \right)}}{{\left( {\sqrt x + 4} \right)\left( {\sqrt x - 4} \right)}}.\dfrac{{\sqrt x + 2}}{{x + 16}}\\
= \dfrac{{x - 4\sqrt x + 4\sqrt x + 16}}{{x - 16}}.\dfrac{{\sqrt x + 2}}{{x + 16}}\\
= \dfrac{{x + 16}}{{x - 16}}.\dfrac{{\sqrt x + 2}}{{x + 16}}\\
= \dfrac{{\sqrt x + 2}}{{x - 16}}\\
3,\\
C = B\left( {A - 1} \right) = \dfrac{{\sqrt x + 2}}{{x - 16}}.\left( {\dfrac{{\sqrt x + 4}}{{\sqrt x + 2}} - 1} \right)\\
= \dfrac{{\sqrt x + 2}}{{x - 16}}.\dfrac{{\left( {\sqrt x + 4} \right) - \left( {\sqrt x + 2} \right)}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2}}{{x - 16}}.\dfrac{2}{{\sqrt x + 2}} = \dfrac{2}{{x - 16}}\\
C \in Z \Leftrightarrow \dfrac{2}{{x - 16}} \in Z \Leftrightarrow x - 16 \in \left\{ { \pm 1;\,\, \pm 2} \right\}\\
\Rightarrow x \in \left\{ {14;\,\,15;\,\,17;\,\,18} \right\}
\end{array}\)
