Cho biểu thức \(A = \frac{{x + 1}}{{x - 1}} - \frac{{x - 1}}{{x + 1}} + \frac{4}{{1 - {x^2}}}\) (\(x \ne 1\); \(x \ne  - 1\)).

Đáp án: \(4\)

Với \(x \ne 1\); \(x \ne - 1\), ta có:

\(A = \frac{{x + 1}}{{x - 1}} - \frac{{x - 1}}{{x + 1}} + \frac{4}{{1 - {x^2}}}\)

\( = \frac{{\left( {x + 1} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \frac{{\left( {x - 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \frac{4}{{\left( {x + 1} \right)\left( {x + 1} \right)}}\)

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

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

\( = \frac{{4x - 4}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{{4\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{4}{{x + 1}}\).

Do đó, \(A = \frac{4}{{x + 1}}\).

Vậy \(a = 4\).