Cho các số nguyên \({\rm{a}},{\rm{b}},{\rm{c}}\) thỏa mãn \(a + \frac{{b + {\rm{lo}}{{\rm{g}}_2}5}}{{c + {\rm{lo}}{{\rm{g}}_2}3}} = {\rm{lo}}{{\rm{g}}_6}45\). Tởng \(a + b + c\) bằng

Phương pháp giải

Sử dụng các công thức:
\({\rm{lo}}{{\rm{g}}_a}b = \frac{{{\rm{lo}}{{\rm{g}}_c}b}}{{{\rm{lo}}{{\rm{g}}_c}a}}\left( {0\left\langle {a,c \ne 1,b} \right\rangle 0} \right)\)\({\rm{lo}}{{\rm{g}}_a}\left( {xy} \right) = {\rm{lo}}{{\rm{g}}_a}x + {\rm{lo}}{{\rm{g}}_a}y\left( {0\left\langle {a \ne 1,x,y} \right\rangle 0} \right)\)\({\rm{lo}}{{\rm{g}}_{{a^n}}}{b^m} = \frac{m}{n}{\rm{lo}}{{\rm{g}}_a}b\left( {0\left\langle {a \ne 1,b} \right\rangle 0} \right)\)

Giải chi tiết

Ta có:
\(a + \frac{{b + {\rm{lo}}{{\rm{g}}_2}5}}{{c + {\rm{lo}}{{\rm{g}}_2}3}} = {\rm{lo}}{{\rm{g}}_6}45 \Leftrightarrow a + \frac{{b + {\rm{lo}}{{\rm{g}}_2}5}}{{c + {\rm{lo}}{{\rm{g}}_2}3}} = \frac{{{\rm{lo}}{{\rm{g}}_2}45}}{{{\rm{lo}}{{\rm{g}}_2}6}}\)\( \Leftrightarrow {\rm{a}} + \frac{{{\rm{b}} + {\rm{lo}}{{\rm{g}}_2}5}}{{{\rm{c}} + {\rm{lo}}{{\rm{g}}_2}3}} = \frac{{{\rm{lo}}{{\rm{g}}_2}\left( {{3^2} \cdot 5} \right)}}{{{\rm{lo}}{{\rm{g}}_2}\left( {2 \cdot 3} \right)}} \Leftrightarrow {\rm{a}} + \frac{{{\rm{b}} + {\rm{lo}}{{\rm{g}}_2}5}}{{{\rm{c}} + {\rm{lo}}{{\rm{g}}_2}3}} = \frac{{2{\rm{lo}}{{\rm{g}}_2}3 + {\rm{lo}}{{\rm{g}}_2}5}}{{1 + {\rm{lo}}{{\rm{g}}_2}3}}\)\( \Leftrightarrow {\rm{a}} + \frac{{{\rm{b}} + {\rm{lo}}{{\rm{g}}_2}5}}{{{\rm{c}} + {\rm{lo}}{{\rm{g}}_2}3}} = \frac{{2 + 2{\rm{lo}}{{\rm{g}}_2}3 - 2 + {\rm{lo}}{{\rm{g}}_2}5}}{{1 + {\rm{lo}}{{\rm{g}}_2}3}}\)\( \Leftrightarrow {\rm{a}} + \frac{{{\rm{b}} + {\rm{lo}}{{\rm{g}}_2}5}}{{{\rm{c}} + {\rm{lo}}{{\rm{g}}_2}3}} = 2 + \frac{{ - 2 + {\rm{lo}}{{\rm{g}}_2}5}}{{1 + {\rm{lo}}{{\rm{g}}_2}3}}\)Đồng nhất hệ số ta có \(a = 2,b = - 2,c = 1\).
Vậy \(a + b + c = 2 + \left( { - 2} \right) + 1 = 1\).
Đáp án: 1