\(\begin{array}{l}
a)\,\,A = \dfrac{{x + 1}}{{x - 2}} + \dfrac{{x - 1}}{{x + 2}} + \dfrac{{{x^2} + 4x}}{{4 - {x^2}}}\\
A = \dfrac{{x + 1}}{{x - 2}} + \dfrac{{x - 1}}{{x + 2}} - \dfrac{{{x^2} + 4x}}{{{x^2} - 4}}\\
A = \dfrac{{\left( {x + 1} \right)\left( {x + 2} \right) + \left( {x - 1} \right)\left( {x - 2} \right) - \left( {{x^2} + 4x} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
A = \dfrac{{{x^2} + 3x + 2 + {x^2} - 3x + 2 - {x^2} - 4x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
A = \dfrac{{{x^2} - 4x + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{{{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{x - 2}}{{x + 2}}
\end{array}\)
b) Với \(x=4\) thì \(A = \dfrac{{4 - 2}}{{4 + 2}} = \dfrac{2}{6} = \dfrac{1}{3}\).
