Đáp án:
$\begin{array}{l}
1)Dkxd:\left\{ \begin{array}{l}
9 + 4x - {x^2} \ge 0\\
x - 3\# 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{x^2} - 4x - 9 \le 0\\
x\# 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 2} \right)^2} \le 13\\
x\# 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2 - \sqrt {13} \le x \le 2 + \sqrt {13} \\
x\# 3
\end{array} \right.\\
A = 2x\sqrt {9 + 4x - {x^2}} + \dfrac{3}{{x - 3}}\\
A \in Z\\
\Leftrightarrow \dfrac{3}{{x - 3}} \in Z\\
\Leftrightarrow \left( {x - 3} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
\Leftrightarrow x \in \left\{ {0;2;4;6} \right\}\\
Do:2 - \sqrt {13} \le x \le 2 + \sqrt {13} \\
\Leftrightarrow x \in \left\{ {0;2;4} \right\}\\
+ Khi:x = 0\\
\Leftrightarrow A = - 1\left( {tm} \right)\\
+ Khi:x = 2\\
\Leftrightarrow A = 2.2.\sqrt {9 + 4.2 - {2^2}} + \dfrac{3}{{2 - 3}} = 4\sqrt {13} - 1\left( {ktm} \right)\\
+ Khi:x = 4\\
\Leftrightarrow A = 2.4.\sqrt {9 + 4.4 - {4^2}} + \dfrac{3}{{4 - 3}} = 27\left( {tm} \right)\\
Vậy\,x \in \left\{ {0;4} \right\}\\
2)Dkxd:\left\{ \begin{array}{l}
{x^2} - 1 \ge 0\\
5 - {x^2} > 0
\end{array} \right. \Leftrightarrow 1 \le {x^2} < 5\\
B = \dfrac{{\sqrt {{x^2} - 1} }}{3} + \dfrac{8}{{\sqrt {4\left( {5 - {x^2}} \right)} }}\\
= \dfrac{{\sqrt {{x^2} - 1} }}{3} + \dfrac{4}{{\sqrt {5 - {x^2}} }}\\
B \in Z\\
\Leftrightarrow \dfrac{4}{{\sqrt {5 - {x^2}} }} \in Z\\
\Leftrightarrow \sqrt {5 - {x^2}} \in \left\{ {1;2;4} \right\}\\
\Leftrightarrow 5 - {x^2} \in \left\{ {1;4;16} \right\}\\
\Leftrightarrow {x^2} \in \left\{ {4;1; - 11} \right\}\\
Do:{x^2} \ge 0\\
\Leftrightarrow {x^2} \in \left\{ {1;4} \right\}\\
\Leftrightarrow x \in \left\{ { - 2; - 1;1;2} \right\}\\
+ Khi:x = - 2;x = 2\\
\Leftrightarrow B = \dfrac{{\sqrt {{x^2} - 1} }}{3} + \dfrac{8}{{\sqrt {4\left( {5 - {x^2}} \right)} }} = \dfrac{{\sqrt 3 }}{3} + 4\left( {ktm} \right)\\
+ Khi:x = - 1;x = 1\\
\Leftrightarrow B = 2\left( {tm} \right)\\
Vậy\,x = - 1;x = 1
\end{array}$
