Cho \(\log 2 = a;log3 = b.\) Tính \({\log_6}90\) theo a, b. A. \(lo{g_6}90 = \frac{{2b - 1}}{{a + b}}\) B. \(lo{g_6}90 = \frac{{b+1}}{{a + b}}\) C. \(lo{g_6}90 = \frac{{2b +1}}{{a + b}}\) D. \(lo{g_6}90 = \frac{{2b + 1}}{{a +2 b}}\)