Cho log^6 (45) = a + (log^2(5+b) / (log ^2 (3+c), a,b,c thuộc Z

* Hướng dẫn giải

Đáp án A

Ta có: ${\log }_{6}45=\frac{{\log }_{2}45}{{\log }_{2}6}=\frac{{\log }_2\left(3^2.5\right)}{{\log }_2\left(2.3\right)}=\frac{2{\log }_{2}3+{\log }_{2}5}{1+{\log }_{2}3}=\frac{2\left(1+{\log }_{2}3\right)+{\log }_{2}5-2}{{\log }_{2}3+1}$

 $=2+\frac{{\log }_{2}5-2}{{\log }_{2}3+1}\Rightarrow \left\{\begin{array}{l}a=2\\ b=-2\\ c=1\end{array}\right.\Rightarrow a+b+c=1.$