Đáp án:

$\begin{array}{l}
a)\dfrac{{4{a^2} + 12a + 9}}{{2{a^2} - a - 6}}\\
 = \dfrac{{{{\left( {2a} \right)}^2} + 2.2a.3 + {3^2}}}{{2{a^2} + 3a - 4a - 6}}\\
 = \dfrac{{{{\left( {2a + 3} \right)}^2}}}{{\left( {2a + 3} \right)\left( {a - 2} \right)}}\\
 = \dfrac{{2a + 3}}{{a - 2}}\\
b)\dfrac{{{x^2} - xy + 2x - 2y}}{{{x^2} - {y^2} + x - y}}\\
 = \dfrac{{x\left( {x - y} \right) + 2\left( {x - y} \right)}}{{\left( {x - y} \right)\left( {x + y + 1} \right)}}\\
 = \dfrac{{\left( {x - y} \right)\left( {x + 2} \right)}}{{\left( {x - y} \right)\left( {x + y + 1} \right)}}\\
 = \dfrac{{x + 2}}{{x + y + 1}}\\
c)\dfrac{{3{x^3} - 7{x^2} + 5x - 1}}{{2{x^3} - {x^2} - 4x + 3}}\\
 = \dfrac{{3{x^3} - 3{x^2} - 4{x^2} + 4x + x - 1}}{{2{x^3} - 2{x^2} + {x^2} - x - 3x + 3}}\\
 = \dfrac{{\left( {x - 1} \right)\left( {3{x^2} - 4x + 1} \right)}}{{\left( {x - 1} \right)\left( {2{x^2} + x - 3} \right)}}\\
 = \dfrac{{3{x^2} - 3x - x + 1}}{{2{x^2} + 3x - 2x - 3}}\\
 = \dfrac{{\left( {x - 1} \right)\left( {3x - 1} \right)}}{{\left( {2x + 3} \right)\left( {x - 1} \right)}}\\
 = \dfrac{{3x - 1}}{{2x + 3}}\\
d)\dfrac{{{a^4} - 3{a^2} + 1}}{{{a^4} - {a^2} - 2a - 1}}\\
 = \dfrac{{{a^4} - 2{a^2} + 1 - {a^2}}}{{{a^4} - \left( {{a^2} + 2a + 1} \right)}}\\
 = \dfrac{{{{\left( {{a^2} - 1} \right)}^2} - {a^2}}}{{{{\left( {{a^2}} \right)}^2} - {{\left( {a + 1} \right)}^2}}}\\
 = \dfrac{{\left( {{a^2} - 1 - a} \right)\left( {{a^2} - 1 + a} \right)}}{{\left( {{a^2} - a - 1} \right)\left( {{a^2} + a + 1} \right)}}\\
 = \dfrac{{{a^2} + a - 1}}{{{a^2} + a + 1}}
\end{array}$