Cho \[a,{\rm{ }}b,{\rm{ }}c\] đều khác 0 thỏa mãn \({4^a} = {25^b} = {10^c}.\) Tính \(T = \frac{c}{a} + \frac{c}{b}.\)

Đặt \({4^a} = {25^b} = {10^c} = t.\)

• \({4^a} = t \Leftrightarrow a = {\log _4}t\);                   

• \({25^b} = t \Leftrightarrow b = {\log _5}t\);                                   

• \({10^c} = t \Leftrightarrow c = {\log _{10}}t\).

Suy ra \(T = \frac{c}{a} + \frac{c}{b} = \frac{{{{\log }_{10}}t}}{{{{\log }_4}t}} + \frac{{{{\log }_{10}}t}}{{{{\log }_{25}}t}} = \frac{{{{\log }_t}4}}{{{{\log }_t}10}} + \frac{{{{\log }_t}25}}{{{{\log }_t}10}}\)\( = \frac{{{{\log }_t}4.25}}{{{{\log }_t}10}} = \frac{{{{\log }_t}100}}{{{{\log }_t}10}} = 2.\)

Chọn C.