Đáp án: A=(81^1/4-1/2log9 4 +20^log125 8) - Đáp án (l)A = (81^((1/4))) - (1/2/log _9)4 + (20^(((log )_(125))8)) = √[4]((81)) - (1/2)(log _((3^2))/2^2) + (20^(((log )_((5^3))/2^3))) = 3 - (1/2)(log _3)2 + (20^(((log )_5)2)) = 3 - (1/2)(log _3)2 + (2^(((log )_5)20)) = 3 - (1/2)(log _3)2 + (2^(((log )_5)( ((2^2)5) ))) = 3 - (1/2)(log _3)2 + (2^(...

Đáp án (l)A = (81^((1/4))) - (1/2/log _9)4 + (20^(((log )_(125))8)) = √[4]((81)) - (1/2)(log _((3^2))/2^2) + (20^(((log )_((5^3))/2^3))) = 3 - (1/2)(log _3)2 + (20^(((log )_5)2)) = 3 - (1/2)(log _3)2 + (2^(((log )_5)20)) = 3 - (1/2)(log _3)2 + (2^(((log )_5)( ((2^2)5) ))) = 3 - (1/2)(log _3)2 + (2^(...

Nhi'xx Nhi'ss

2 năm trước

A=(81^1/4-1/2log9 4 +20^log125 8)

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maitran15

2 năm trước

Đáp án:

$\begin{array}{l}
A = {81^{\frac{1}{4}}} - \frac{1}{2}{\log _9}4 + {20^{{{\log }_{125}}8}}\\
= \sqrt[4]{{81}} - \frac{1}{2}.{\log _{{3^2}}}{.2^2} + {20^{{{\log }_{{5^3}}}{2^3}}}\\
= 3 - \frac{1}{2}.{\log _3}2 + {20^{{{\log }_5}2}}\\
= 3 - \frac{1}{2}.{\log _3}2 + {2^{{{\log }_5}20}}\\
= 3 - \frac{1}{2}.{\log _3}2 + {2^{{{\log }_5}\left( {{2^2}.5} \right)}}\\
= 3 - \frac{1}{2}.{\log _3}2 + {2^{1 + 2{{\log }_5}2}}\\
= 3 - \frac{1}{2}.{\log _3}2 + 2.1,816\\
= 6,318
\end{array}$

(Lần sau bạn ghi rõ đề ra nhé)