Đáp án:

$\begin{array}{l}
a)2{x^2} + 7x - 15\\
 = 2{x^2} + 10x - 3x - 15\\
 = \left( {x + 5} \right)\left( {2x - 3} \right)\\
{x^2} + 3x - 10\\
 = \left( {x - 2} \right)\left( {x + 5} \right)\\
 \Rightarrow MSC = \left( {x + 5} \right)\left( {2x - 3} \right)\left( {x - 2} \right)\\
 \Rightarrow \left\{ \begin{array}{l}
\dfrac{x}{{2{x^2} + 7x - 15}} = \dfrac{{x\left( {x - 2} \right)}}{{\left( {x + 5} \right)\left( {2x - 3} \right)\left( {x - 2} \right)}}\\
 = \dfrac{{{x^2} - 2x}}{{\left( {x + 5} \right)\left( {2x - 3} \right)\left( {x - 2} \right)}}\\
\dfrac{{x + 2}}{{{x^2} + 3x - 10}} = \dfrac{{\left( {x + 2} \right)\left( {2x - 3} \right)}}{{\left( {x + 5} \right)\left( {2x - 3} \right)\left( {x - 2} \right)}}\\
 = \dfrac{{2{x^2} + x - 6}}{{\left( {x + 5} \right)\left( {2x - 3} \right)\left( {x - 2} \right)}}\\
\dfrac{1}{{x + 5}} = \dfrac{{\left( {2x - 3} \right)\left( {x - 2} \right)}}{{\left( {x + 5} \right)\left( {2x - 3} \right)\left( {x - 2} \right)}} = \dfrac{{2{x^2} - 7x + 6}}{{\left( {x + 5} \right)\left( {2x - 3} \right)\left( {x - 2} \right)}}
\end{array} \right.\\
b) - {x^2} + 3x - 2\\
 =  - \left( {x - 1} \right)\left( {x - 2} \right)\\
{x^2} + 5x - 6 = \left( {x - 1} \right)\left( {x + 6} \right)\\
 - {x^2} + 4x - 3 =  - \left( {x - 1} \right)\left( {x - 3} \right)\\
 \Rightarrow MSC = \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x + 6} \right)\\
 \Rightarrow  + )\dfrac{1}{{ - {x^2} + 3x - 2}} = \dfrac{{ - \left( {x - 3} \right)\left( {x + 6} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x + 6} \right)}}\\
 + )\dfrac{1}{{{x^2} + 5x - 6}} = \dfrac{{ - \left( {x - 2} \right)\left( {x - 3} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x + 6} \right)}}\\
 + )\dfrac{1}{{ - {x^2} + 4x - 3}} = \dfrac{{ - \left( {x - 2} \right)\left( {x + 6} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x + 6} \right)}}\\
c)MSC:\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
\dfrac{3}{{{x^3} - 1}}\\
\dfrac{{2x}}{{{x^2} + x + 1}} = \dfrac{{2x\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
\dfrac{x}{{x - 1}} = \dfrac{{x\left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
d)\\
 + ){x^2} - 2xy + {y^2} - {z^2} = {\left( {x - y} \right)^2} - {z^2}\\
 = \left( {x - y - z} \right)\left( {x - y + z} \right)\\
 + ){x^2} + 2yz + {y^2} - {z^2}\\
 = {x^2} - {\left( {y - z} \right)^2}\\
 = \left( {x - y + z} \right)\left( {x + y - z} \right)\\
 + ){x^2} - 2xz - {y^2} + {z^2}\\
 = {\left( {x - z} \right)^2} - {y^2}\\
 = \left( {x - y - z} \right)\left( {x + y - z} \right)\\
 \Rightarrow MSC = \left( {x - y - z} \right)\left( {x - y + z} \right)\left( {x + y - z} \right)
\end{array}$