Giải thích các bước giải:
ĐKXĐ: \(x \ne \pm 2\)
Ta có:
\(\begin{array}{l}
A = \dfrac{1}{{x - 2}} + \dfrac{{2x + 8}}{{{x^2} - 4}} + \dfrac{1}{{x + 2}}\\
= \dfrac{1}{{x - 2}} + \dfrac{{2x + 8}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} + \dfrac{1}{{x + 2}}\\
= \dfrac{{\left( {x + 2} \right) + 2x + 8 + \left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{4x + 8}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{4\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{4}{{x - 2}}\\
A = 2 \Leftrightarrow \dfrac{4}{{x - 2}} = 2 \Leftrightarrow x - 2 = 2 \Leftrightarrow x = 4
\end{array}\)
