Đáp án:
$\begin{array}{l}
1)y = \dfrac{{\sqrt x + 1}}{{\sqrt x + 2}} = \dfrac{{\sqrt x + 2 - 1}}{{\sqrt x + 2}} = 1 - \dfrac{1}{{\sqrt x + 2}}\\
y \in Z\\
\Rightarrow \dfrac{1}{{\sqrt x + 2}} \in Z\\
\Rightarrow \left( {\sqrt x + 2} \right) \in I\left( 1 \right)\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x + 2 = 1\\
\sqrt x + 2 = - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = - 1\left( {ktm} \right)\\
\sqrt x = - 3\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow x \in \emptyset \\
2)y = \dfrac{{2\sqrt x }}{{\sqrt x - 1}} = \dfrac{{2\sqrt x - 2 + 2}}{{\sqrt x - 1}}\\
= 2 + \dfrac{2}{{\sqrt x - 1}}\\
y \in Z\\
\Rightarrow \dfrac{2}{{\sqrt x - 1}} \in Z\\
\Rightarrow \left( {\sqrt x - 1} \right) \in \left\{ { - 1;1;2} \right\}\left( {do:\sqrt x - 1 \ge - 1} \right)\\
\Rightarrow \sqrt x \in \left\{ {0;2;3} \right\}\\
\Rightarrow x \in \left\{ {0;4;9} \right\}\\
4)y = \dfrac{{\sqrt x - 1}}{{\sqrt x - 2}} = \dfrac{{\sqrt x - 2 + 1}}{{\sqrt x - 2}} = 1 + \dfrac{1}{{\sqrt x - 2}}\\
\Rightarrow \dfrac{1}{{\sqrt x - 2}} \in Z\\
\Rightarrow \left( {\sqrt x - 2} \right) \in \left\{ { - 1;1} \right\}\\
\Rightarrow \sqrt x \in \left\{ {1;3} \right\}\\
\Rightarrow x \in \left\{ {1;9} \right\}\\
5)y = \dfrac{{8\sqrt x }}{{x + \sqrt x + 4}}\\
\Rightarrow y.x + y.\sqrt x + 4y = 8\sqrt x \\
\Rightarrow y.x + \left( {y - 8} \right).\sqrt x + 4y = 0\\
+ Khi:y = 0 \Rightarrow x = 0\left( {tmdk} \right)\\
+ Khi:y \ne 0\\
\Rightarrow \left\{ \begin{array}{l}
\Delta \ge 0\\
\dfrac{{8 - y}}{y} \ge 0\\
4 \ge 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{\left( {y - 8} \right)^2} - 4.y.4y \ge 0\\
0 < y \le 8
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- 15{y^2} - 16y + 64 \ge 0\\
0 < y \le 8
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\left( {3y + 8} \right)\left( {5y - 8} \right) \le 0\\
0 < y \le 8
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
- \dfrac{8}{3} \le y \le \dfrac{8}{5}\\
0 < y \le 8
\end{array} \right.\\
\Rightarrow 0 < y \le \dfrac{8}{5}\\
y \in Z\\
\Rightarrow y = 1\\
\Rightarrow x - 7\sqrt x + 4 = 0\\
\Rightarrow \sqrt x = \dfrac{{7 \pm \sqrt {33} }}{2}\left( {ktm:do:x \in R} \right)\\
Vậy\,x = 0
\end{array}$
