Cho 2 số thực dương a , b thỏa mãn √ a ≠ b , a ≠ 1 , log a/b = 2 . Tính T = log √ a b 3 √ ba .
Giải chi tiết
Ta có: \({\rm{lo}}{{\rm{g}}_a}b = 2 \Rightarrow {\rm{lo}}{{\rm{g}}_b}a = \frac{1}{2}\).
\(T = {\rm{lo}}{{\rm{g}}_{\frac{{\sqrt a }}{b}}}\sqrt[3]{{ba}} = {\rm{lo}}{{\rm{g}}_{\frac{{\sqrt a }}{b}}}\sqrt[3]{b} + {\rm{lo}}{{\rm{g}}_{\frac{{\sqrt a }}{b}}}\sqrt[3]{a}\).
\( = \frac{1}{{{\rm{lo}}{{\rm{g}}_{\sqrt[3]{b}}}\frac{{\sqrt a }}{b}}} + \frac{1}{{{\rm{lo}}{{\rm{g}}_{\sqrt[3]{a}}}\frac{{\sqrt a }}{b}}}\).
\( = \frac{1}{{{\rm{lo}}{{\rm{g}}_{\sqrt[3]{b}}}\sqrt a - {\rm{lo}}{{\rm{g}}_{\sqrt[3]{b}}}b}} + \frac{1}{{{\rm{lo}}{{\rm{g}}_{\sqrt[3]{a}}}\sqrt a - {\rm{lo}}{{\rm{g}}_{\sqrt[3]{a}}}b}}\).
\( = \frac{1}{{\frac{3}{2}{\rm{lo}}{{\rm{g}}_b}a - 3}} + \frac{1}{{\frac{3}{2} - 3{\rm{lo}}{{\rm{g}}_a}b}}\).
\( = \frac{1}{{\frac{3}{2} \cdot \frac{1}{2} - 3}} + \frac{1}{{\frac{3}{2} - 3.2}} = - \frac{2}{3}\).
Đáp án cần chọn là: D