Đáp án:
$abc = 48$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
{\log _{40}}75 = {\log _{40}}\left( {{{3.5}^2}} \right)\\
= {\log _{40}}3 + 2{\log _{40}}5\\
= \dfrac{1}{{{{\log }_3}40}} + \dfrac{2}{{{{\log }_5}40}}\\
= \dfrac{1}{{{{\log }_3}\left( {{2^3}.5} \right)}} + \dfrac{2}{{{{\log }_5}\left( {{2^3}.5} \right)}}\\
= \dfrac{1}{{3{{\log }_3}2 + {{\log }_3}5}} + \dfrac{2}{{3{{\log }_5}2 + 1}}\\
= \dfrac{1}{{{{\log }_3}2\left( {3 + {{\log }_2}5} \right)}} + \dfrac{2}{{3.\dfrac{1}{{{{\log }_2}5}} + 1}}\\
= \dfrac{1}{{{{\log }_3}2}}.\dfrac{1}{{3 + {{\log }_2}5}} + \dfrac{{2{{\log }_2}5}}{{3 + {{\log }_2}5}}\\
= \dfrac{{{{\log }_2}3}}{{3 + {{\log }_2}5}} + \dfrac{{2{{\log }_2}5}}{{3 + {{\log }_2}5}}\\
= \dfrac{{{{\log }_2}3}}{{3 + {{\log }_2}5}} + \dfrac{{2\left( {3 + {{\log }_2}5} \right) - 6}}{{3 + {{\log }_2}5}}\\
= 2 + \dfrac{{{{\log }_2}3 - 6}}{{3 + {{\log }_2}5}}
\end{array}$
Để ${\log _{40}}75 = a + \dfrac{{{{\log }_2}3 - b}}{{c + {{\log }_2}5}}$
$\begin{array}{l}
\Leftrightarrow 2 + \dfrac{{{{\log }_2}3 - 6}}{{3 + {{\log }_2}5}} = a + \dfrac{{{{\log }_2}3 - b}}{{c + {{\log }_2}5}}\\
\Leftrightarrow \left\{ \begin{array}{l}
a = 2\\
b = 6\\
c = 3
\end{array} \right.\\
\Rightarrow abc = 2.6.3 = 48
\end{array}$
Vậy $abc = 48$
