\(A = \frac{a}{{ab + a + 2}} + \frac{b}{{bc + b + 1}} + \frac{{2c}}{{ac + 2c + 2}}\)

\( = \frac{a}{{ab + a + 2}} + \frac{{ab}}{{a\left( {bc + b + 1} \right)}} + \frac{{2abc}}{{ab\left( {ac + 2c + 2} \right)}}\)

\( = \frac{a}{{ab + a + 2}} + \frac{{ab}}{{abc + ab + a}} + \frac{{2abc}}{{{a^2}bc + 2abc + 2ab}}\)

Mà abc = 2 nên \(A = \frac{a}{{ab + a + 2}} + \frac{{ab}}{{2 + ab + a}} + \frac{{2.2}}{{a.2 + 2.2 + 2a}}\)

\( = \frac{{a + ab + 2}}{{ab + a + 2}}\)= 1.