Tính S = m^ 2 + n^ 2 .

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Cho \({\rm{lo}}{{\rm{g}}_2}5 = a,{\rm{lo}}{{\rm{g}}_5}3 = b\), biết \({\rm{lo}}{{\rm{g}}_{24}}15 = \frac{{ma + ab}}{{n + ab}}\), với m, n thuộc R. Tính \(S = {m^2} + {n^2}\).

\(S = 2\).

\(S = 10\).

\(S = 5\).

\(S = 13\).

Giải thích

Đáp án B

Hướng dẫn giải

Ta có: \({\rm{lo}}{{\rm{g}}_3}2 = {\rm{lo}}{{\rm{g}}_3}5.{\rm{lo}}{{\rm{g}}_5}2 = \frac{1}{{{\rm{lo}}{{\rm{g}}_2}5.{\rm{lo}}{{\rm{g}}_5}3}} = \frac{1}{{ab}};{\rm{lo}}{{\rm{g}}_5}2 = \frac{1}{{{\rm{lo}}{{\rm{g}}_2}5}} = \frac{1}{a}.\)

Do đó: \({\rm{lo}}{{\rm{g}}_{24}}15 = {\rm{lo}}{{\rm{g}}_{24}}3 + {\rm{lo}}{{\rm{g}}_{24}}5 = \frac{1}{{{\rm{lo}}{{\rm{g}}_3}24}} + \frac{1}{{{\rm{lo}}{{\rm{g}}_5}24}} = \frac{1}{{1 + {\rm{lo}}{{\rm{g}}_3}8}} + \frac{1}{{{\rm{lo}}{{\rm{g}}_5}3 + {\rm{lo}}{{\rm{g}}_5}8}}\).

\( = \frac{1}{{1 + 3{\rm{lo}}{{\rm{g}}_3}2}} + \frac{1}{{{\rm{lo}}{{\rm{g}}_5}3 + 3{\rm{lo}}{{\rm{g}}_5}2}} = \frac{1}{{1 + \frac{3}{{ab}}}} + \frac{1}{{b + \frac{3}{a}}} = \frac{{ab + a}}{{ab + 3}}\)

\( \Rightarrow m = 1,n = 3 \Rightarrow S = 10\).