Cho \(a=\log_23\), \(b=\log_53\). Số \(\log_645\) bằng
\(\dfrac{a+2ab}{ab}\).\(\dfrac{2a^2-2ab}{ab}\).\(\dfrac{a+2ab}{ab+b}\).\(\dfrac{2a^2-2ab}{ab+b}\).Hướng dẫn giải:\(\log_645=\dfrac{\log_345}{\log_36}=\dfrac{\log_3\left(3^2.5\right)}{\log_3\left(2.3\right)}\)
\(=\dfrac{2+\log_35}{\log_32+1}=\dfrac{2+\dfrac{1}{b}}{\dfrac{1}{a}+1}=\dfrac{a+2ab}{ab+b}\).