Đáp án:
A
Giải thích các bước giải:
\(\begin{array}{l}
\log _4^2\frac{{125}}{{81}} = {({\log _4}125 - {\log _4}81)^2} = \log _{{2^2}}^2{5^3} + \log _{{2^2}}^2{3^4} - 2{\log _{{2^2}}}{5^3}.{\log _{{2^2}}}{3^4}\\
= {(3.\frac{1}{2}{\log _2}5)^2} + {(4.\frac{1}{2}{\log _2}3)^2} - 2.\frac{1}{2}.3.\frac{1}{2}.4{\log _2}5.{\log _2}3 = \frac{9}{4}{b^2} + 4{a^2} - 6ab\\
- > m = \frac{9}{4},n = 4,k = - 6\\
- > 4m - n + 2k = - 7
\end{array}\)
