Đáp án:
$\begin{array}{l}
\dfrac{{5x - 2}}{{{x^2} + x - 20}} = \dfrac{a}{{x + 5}} - \dfrac{b}{{x - 4}}\\
\Rightarrow \dfrac{{5x - 2}}{{{x^2} + 5x - 4x - 20}} = \dfrac{{a\left( {x - 4} \right) - b\left( {x + 5} \right)}}{{\left( {x + 5} \right)\left( {x - 4} \right)}}\\
\Rightarrow \dfrac{{5x - 2}}{{\left( {x + 5} \right)\left( {x - 4} \right)}} = \dfrac{{a.x - 4a - b.x - 5b}}{{\left( {x + 5} \right)\left( {x - 4} \right)}}\\
\Rightarrow 5x - 2 = \left( {a - b} \right).x - 4a - 5b\\
\Rightarrow \left\{ \begin{array}{l}
5 = a - b\\
- 2 = - 4a - 5b
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a - b = 5\\
4a + 5b = 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
4a - 4b = 20\\
4a + 5b = 2
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
9b = - 18\\
a = 5 + b
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
b = - 2\\
a = 3
\end{array} \right.\\
Vậy\,a = 3;b = - 2
\end{array}$
