A=(81^1/4-1/2log9 4 +25^log125 8).49^log7 2 21/08/2021 Bởi Genesis A=(81^1/4-1/2log9 4 +25^log125 8).49^log7 2
Đáp án: $\begin{array}{l}A = \left( {{{81}^{\frac{1}{4}}} – \frac{1}{2}{{\log }_9}4 + {{25}^{{{\log }_{125}}8}}} \right){.49^{{{\log }_7}2}}\\ = \left( {\sqrt[4]{{81}} – \frac{1}{2}{{\log }_3}2 + {{25}^{{{\log }_5}2}}} \right){.7^{2.{{\log }_7}2}}\\ = \left( {3 – \frac{1}{2}{{\log }_3}2 + {{\left( {{5^{{{\log }_5}2}}} \right)}^2}} \right).{\left( {{7^{{{\log }_7}2}}} \right)^2}\\ = \left( {3 – \frac{1}{2}{{\log }_3}2 + {{\left( {{2^{{{\log }_5}5}}} \right)}^2}} \right).{\left( {{2^{{{\log }_7}7}}} \right)^2}\\ = \left( {3 – \frac{1}{2}{{\log }_3}2 + 4} \right).4\\ = 28 – 2{\log _3}2\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
A = \left( {{{81}^{\frac{1}{4}}} – \frac{1}{2}{{\log }_9}4 + {{25}^{{{\log }_{125}}8}}} \right){.49^{{{\log }_7}2}}\\
= \left( {\sqrt[4]{{81}} – \frac{1}{2}{{\log }_3}2 + {{25}^{{{\log }_5}2}}} \right){.7^{2.{{\log }_7}2}}\\
= \left( {3 – \frac{1}{2}{{\log }_3}2 + {{\left( {{5^{{{\log }_5}2}}} \right)}^2}} \right).{\left( {{7^{{{\log }_7}2}}} \right)^2}\\
= \left( {3 – \frac{1}{2}{{\log }_3}2 + {{\left( {{2^{{{\log }_5}5}}} \right)}^2}} \right).{\left( {{2^{{{\log }_7}7}}} \right)^2}\\
= \left( {3 – \frac{1}{2}{{\log }_3}2 + 4} \right).4\\
= 28 – 2{\log _3}2
\end{array}$