$\begin{array}{l} \dfrac{1}{{{x^2} + x}} + \dfrac{1}{{{x^2} + 3x + 2}} + \dfrac{1}{{{x^2} + 5x + 6}} + \dfrac{1}{{{x^2} + 7x + 12}} + \dfrac{1}{{{x^2} + 9x + 20}}\\ = \dfrac{1}{{x\left( {x + 1} \right)}} + \dfrac{1}{{\left( {x + 1} \right)\left( {x + 2} \right)}} + \dfrac{1}{{\left( {x + 2} \right)\left( {x + 3} \right)}} + \dfrac{1}{{\left( {x + 3} \right)\left( {x + 4} \right)}} + \dfrac{1}{{\left( {x + 4} \right)\left( {x + 5} \right)}}\\ = \dfrac{1}{x} - \dfrac{1}{{x + 1}} + \dfrac{1}{{x + 1}} - \dfrac{1}{{x + 2}} + \dfrac{1}{{x + 2}} - \dfrac{1}{{x + 3}} + \dfrac{1}{{x + 3}} - \dfrac{1}{{x + 4}} + \dfrac{1}{{x + 4}} - \dfrac{1}{{x + 5}}\\ = \dfrac{1}{x} - \dfrac{1}{{x + 5}} = \dfrac{{x + 5 - x}}{{x\left( {x + 5} \right)}} = \dfrac{5}{{x\left( {x + 5} \right)}}\\ \left( {\dfrac{3}{{a - 3}} - \dfrac{{2a}}{{9 - {a^2}}} + \dfrac{a}{{a + 3}}} \right):\dfrac{{2a}}{{a + 3}}\\ = \left( {\dfrac{{3\left( {a + 3} \right) + 2a + a\left( {a - 3} \right)}}{{\left( {a - 3} \right)\left( {a + 3} \right)}}} \right).\dfrac{{a + 3}}{{2a}}\\ = \dfrac{{{a^2} + 2a + 9}}{{2a\left( {a - 3} \right)}} \end{array}$
