a. ĐK: x ≠ ± 2

\(A = \frac{{\left( {\frac{{2 - 3x}}{{{x^2} - 4}} - \frac{2}{{x - 2}} - \frac{1}{{x + 2}}} \right)}}{{\left( {x - 2 + \frac{{10 - {x^2}}}{{x + 2}}} \right)}} = \frac{{\frac{{2 - 3x - 2\left( {x + 2} \right) - \left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}}}{{\frac{{\left( {x - 2} \right)\left( {x + 2} \right) + 10 - {x^2}}}{{x + 2}}}}\)

\(A = \frac{{\frac{{2 - 3x - 2x - 4 - x + 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}}}{{\frac{{{x^2} - 4 + 10 - {x^2}}}{{x + 2}}}} = \frac{{ - 6x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\frac{{x + 2}}{6} = \frac{{ - x}}{{x - 2}} = \frac{x}{{2 - x}}\)

b. \(A = \frac{x}{{2 - x}}\)

\(\left| x \right| = \frac{1}{2} \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{1}{2}}\\{x = - \frac{1}{2}}\end{array}} \right.\)

TH1: \(x = \frac{1}{2}\), ta có: \(A = \frac{{\frac{1}{2}}}{{2 - \frac{1}{2}}} = \frac{{\frac{1}{2}}}{{\frac{3}{2}}} = \frac{1}{3}\)

TH2: \(x = - \frac{1}{2}\), ta có:

\(A = \frac{{ - \frac{1}{2}}}{{2 + \frac{1}{2}}} = \frac{{ - \frac{1}{2}}}{{\frac{5}{2}}} = - \frac{1}{5}\).