Đáp án:
$\begin{array}{l}
A = \dfrac{{x - 1}}{{x + 2}} + \dfrac{{5x - 2}}{{{x^2} - 4}}\\
= \dfrac{{\left( {x - 1} \right)\left( {x - 2} \right) + 5x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} - 3x + 2 + 5x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} + 2x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{x\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{x}{{x - 2}}\\
B = \dfrac{{4{x^2}}}{{{x^2} - 9}} - \dfrac{{x - 3}}{{x + 3}} + \dfrac{{x + 3}}{{x - 3}}\\
= \dfrac{{4{x^2} - {{\left( {x - 3} \right)}^2} + {{\left( {x + 3} \right)}^2}}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{4{x^2} - {x^2} + 6x - 9 + {x^2} + 6x + 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{4{x^2} + 12x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{4x}}{{x - 3}}\\
C = \left( {\dfrac{{x - 1}}{x} + \dfrac{{2x + 1}}{{{x^2} + x}}} \right).\dfrac{x}{{x + 2}}\\
= \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right) + 2x + 1}}{{x\left( {x + 1} \right)}}.\dfrac{x}{{x + 2}}\\
= \dfrac{{{x^2} - 1 + 2x + 1}}{{x + 1}}.\dfrac{1}{{x + 2}}\\
= \dfrac{x}{{x + 1}}\\
D = \dfrac{{{x^2} - 5x - 18}}{{{x^2} + 2x - 15}} + \dfrac{3}{{x - 3}} - \dfrac{2}{{x + 5}}\\
= \dfrac{{{x^2} - 5x - 18 + 3\left( {x + 5} \right) - 2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 5} \right)}}\\
= \dfrac{{{x^2} - 5x - 18 + x + 15 + 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} - 4x + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{\left( {x - 3} \right)\left( {x - 1} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{x - 1}}{{x + 3}}
\end{array}$
