Đáp án:

$\begin{array}{l}
2)A = \left( {\frac{{x - 3}}{x} - \frac{x}{{x - 3}} + \frac{9}{{{x^2} - 3x}}} \right):\frac{{2x - 2}}{x}\\
a)ĐKxđ:\left\{ \begin{array}{l}
x \ne 0\\
x \ne 3\\
x \ne 1
\end{array} \right.\\
A = \left( {\frac{{x - 3}}{x} - \frac{x}{{x - 3}} + \frac{9}{{x\left( {x - 3} \right)}}} \right).\frac{x}{{2x - 2}}\\
 = \frac{{\left( {x - 3} \right)\left( {x - 3} \right) - x.x + 9}}{{x\left( {x - 3} \right)}}.\frac{x}{{2\left( {x - 1} \right)}}\\
 = \frac{{{x^2} - 6x + 9 - {x^2} + 9}}{{x - 3}}.\frac{1}{{2\left( {x - 1} \right)}}\\
 = \frac{{18 - 6x}}{{2\left( {x - 1} \right)\left( {x - 3} \right)}}\\
 = \frac{{ - 6\left( {x - 3} \right)}}{{2\left( {x - 1} \right)\left( {x - 3} \right)}}\\
 = \frac{{ - 3}}{{x - 1}}\\
b)A = 2\\
 \Rightarrow \frac{{ - 3}}{{x - 1}} = 2\\
 \Rightarrow x - 1 =  - \frac{3}{2}\\
 \Rightarrow x = 1 - \frac{3}{2} =  - \frac{1}{2}\left( {tm} \right)\\
Vậy\,x =  - \frac{1}{2}\,thì\,A = 2\\
c)x \ne 0;x \ne 3;x \ne 1\\
A = \frac{{ - 3}}{{x - 1}}\\
A \in Z\\
 \Rightarrow \left( {x - 1} \right) \in U\left( 3 \right) = {\rm{\{ }} - 3; - 1;1;3\} \\
 \Rightarrow x \in {\rm{\{ }} - 2;0;2;4\} \\
Do:x \ne 0;x \ne 3;x \ne 1\\
 \Rightarrow x \in {\rm{\{ }} - 2;2;4\} 
\end{array}$

$\begin{array}{l}
4)5{x^2} + 5{y^2} + 8xy - 2x + 2y + 2 = 0\\
 \Rightarrow 4{x^2} + 4{y^2} + 8xy + {x^2} - 2x + {y^2} + 2y + 2 = 0\\
 \Rightarrow 4\left( {{x^2} + {y^2} + 2xy} \right) + \left( {{x^2} - 2x + 1} \right) + \left( {{y^2} + 2y + 1} \right) = 0\\
 \Rightarrow 4{\left( {x + y} \right)^2} + {\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 0\\
Do:\left\{ \begin{array}{l}
{\left( {x + y} \right)^2} \ge 0\forall x,y\\
{\left( {x - 1} \right)^2} \ge 0\forall x\\
{\left( {y + 1} \right)^2} \ge 0\forall y
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + y = 0\\
x - 1 = 0\\
y + 1 = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 1\\
y =  - 1
\end{array} \right.\\
M = {\left( {x + y} \right)^{2013}} + {\left( {x - 2} \right)^{2014}} + {\left( {y + 1} \right)^{2015}}\\
 = 0 + {\left( {1 - 2} \right)^{2014}} + 0\\
 = {\left( { - 1} \right)^{2014}}\\
 = 1
\end{array}$