1) $\log_2 24 = \log_2 2^3 + \log_2 3 = 3 + \log_2 3$
$\dfrac{1}{2}\log_2 72 = \dfrac{1}{2}[\log_2 2^3 + \log_2 3^2] = \dfrac{3}{2} + 2log_2 3$
Vậy tử $= \dfrac{9}{2} + 3\log_2 3$
$\log_3 18 = \log_3 3^2 +\log_3 2 = 2 + \log_3 2$
$\dfrac{1}{3}\log_3 72 = \dfrac{1}{3}[\log_3 2^3 + \log_3 3^2] = \dfrac{1}{3}[2 + 3\log_3 2] = \dfrac{2}{3} + \log_3 2$
Vậy mẫu $= \dfrac{8}{3} + 2\log_3 2$
Tự ráp nốt