Đáp án:
$\begin{array}{l}
1)A = \dfrac{2}{{\sqrt x - 3}} \in Z\\
\Rightarrow \left( {\sqrt x - 3} \right) \in U\left( 2 \right) = \{ - 2; - 1;1;2\} \\
\Rightarrow \sqrt x \in \{ 1;2;4;5\} \\
\Rightarrow x \in \{ 1;4;16;25\} \\
B = \dfrac{{\sqrt x - 2}}{{\sqrt x + 3}} = \dfrac{{\sqrt x + 3 - 5}}{{\sqrt x + 3}} = 1 - \dfrac{5}{{\sqrt x + 3}}\\
Do:1 \in Z\\
\Rightarrow \dfrac{5}{{\sqrt x + 3}} \in Z\\
\Rightarrow \left( {\sqrt x + 3} \right) \in \{ - 5; - 1;1;5\} \\
Do:\sqrt x + 3 \ge 3\\
\Rightarrow \sqrt x \in \{ 1;5\} \\
\Rightarrow x \in \{ 1;25\} \\
C = \dfrac{{2\sqrt x - 1}}{{\sqrt x + 3}} = \dfrac{{2\left( {\sqrt x + 3} \right) - 6 - 1}}{{\sqrt x + 3}}\\
= 2 - \dfrac{7}{{\sqrt x + 3}}\\
\Rightarrow \dfrac{7}{{\sqrt x + 3}} \in Z\\
\Rightarrow \left( {\sqrt x + 3} \right) = 7\\
\Rightarrow \sqrt x = 4\\
\Rightarrow x = 16\\
2)B = \dfrac{{\sqrt x + 5}}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 + 4}}{{\sqrt x + 1}}\\
= 1 + \dfrac{4}{{\sqrt x + 1}}\\
Do:1 \in Z\\
\Rightarrow \dfrac{4}{{\sqrt x + 1}} \in Z\\
\Rightarrow \left( {\sqrt x + 1} \right) \in \{ 1;2;4\} \\
\left( {do:\sqrt x + 1 \ge 1} \right)\\
\Rightarrow \sqrt x \in \{ 0;1;3\} \\
\Rightarrow x \in \{ 0;1;9\} \\
E = \dfrac{{x + \sqrt x }}{{x + \sqrt x + 1}}\left( {0 \le x < 1} \right)\\
= \dfrac{{x + \sqrt x + 1 - 1}}{{x + \sqrt x + 1}}\\
= 1 - \dfrac{1}{{x + \sqrt x + 1}}\\
\Rightarrow \left( {x + \sqrt x + 1} \right) \in U\left( 1 \right) = 1\\
\Rightarrow x + \sqrt x = 0\\
\Rightarrow \sqrt x \left( {\sqrt x + 1} \right) = 0\\
\Rightarrow x = 0
\end{array}$
