Đáp án:

$\begin{array}{l}
a)\left( {\dfrac{{x + 2}}{{x + 1}} - \dfrac{{2x}}{{x - 1}}} \right).\dfrac{{3x + 3}}{x} + \dfrac{{4{x^2} + x + 7}}{{{x^2} - x}}\\
 = \dfrac{{\left( {x + 2} \right)\left( {x - 1} \right) - 2x\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}.\dfrac{{3\left( {x + 1} \right)}}{x}\\
 + \dfrac{{4{x^2} + x + 7}}{{x\left( {x - 1} \right)}}\\
 = \dfrac{{{x^2} + x - 2 - 2{x^2} - 2x}}{{x - 1}}.\dfrac{3}{x} + \dfrac{{4{x^2} + x + 7}}{{x\left( {x - 1} \right)}}\\
 = \dfrac{{3.\left( { - {x^2} - x - 2} \right) + 4{x^2} + x + 7}}{{x\left( {x - 1} \right)}}\\
 = \dfrac{{{x^2} - 2x + 1}}{{x\left( {x - 1} \right)}}\\
 = \dfrac{{{{\left( {x - 1} \right)}^2}}}{{x\left( {x - 1} \right)}}\\
 = \dfrac{{x - 1}}{x}\\
b)\left( {\dfrac{{2xy}}{{{x^2} - {y^2}}} + \dfrac{{x - y}}{{2x + 2y}}} \right):\dfrac{{x + y}}{{2x}} + \dfrac{y}{{y - x}}\\
 = \dfrac{{2xy.2 + \left( {x - y} \right).\left( {x - y} \right)}}{{2\left( {x + y} \right)\left( {x - y} \right)}}.\dfrac{{2x}}{{x + y}} - \dfrac{y}{{x - y}}\\
 = \dfrac{{4xy + {x^2} - 2xy + {y^2}}}{{{{\left( {x + y} \right)}^2}.\left( {x - y} \right)}}.x - \dfrac{y}{{x - y}}\\
 = \dfrac{{{x^2} + 2xy + {y^2}}}{{{{\left( {x + y} \right)}^2}.\left( {x - y} \right)}}.x - \dfrac{y}{{x - y}}\\
 = \dfrac{x}{{x - y}} - \dfrac{y}{{x - y}}\\
 = \dfrac{{x - y}}{{x - y}}\\
 = 1
\end{array}$