\(\dfrac{81}{125}=\dfrac{3^4}{5^3}\)

\(\dfrac{-8}{27}=\dfrac{\left(-2\right)^3}{3^3}=\left(\dfrac{-2}{3}\right)^3\)

Ta có:

\(\dfrac{81}{125}=\dfrac{3^4}{5^4}=\left(\dfrac{3}{5}\right)^4\)

\(-\dfrac{8}{27}=-\dfrac{2^3}{3^3}=\left(-\dfrac{2}{3}\right)^3\)

\(\dfrac{81}{125}=\dfrac{3^4}{5^4}=\left(\dfrac{3}{5}\right)^4\)

\(\dfrac{-8}{27}=\dfrac{\left(-2\right)^3}{3^3}=\left(\dfrac{-2}{3}\right)^3\)

Chúc bạn học tốt!

\(a:\dfrac{81}{125}=\dfrac{3^4}{5^3}\)

\(b:\dfrac{-8}{27}=-\dfrac{2^3}{3^3}\)

81/125 bằng 81/125 mũ 1

-8/27 bằng -2/3 mũ 3

câu này mink đã được học rồi nên chắc đúng

0,25 = (0,5)²

\(\dfrac{1}{49}\) = (\(\dfrac{1}{7}\))²

- \(\dfrac{27}{125}\) = (- \(\dfrac{3}{5}\))³

 \(\dfrac{81}{16}\) = ( \(\dfrac{9}{4}\))²

\(\dfrac{169}{196}\) = (\(\dfrac{13}{14}\))²

Bài 2: \(\dfrac{81}{225}=\dfrac{9^2}{15^2}=\left(\dfrac{9}{15}\right)^2\)

\(\left(-\dfrac{8}{27}\right)=\left(-\dfrac{2^3}{3^3}\right)=\left(-\dfrac{2}{3}\right)^3\)

-8/27 = (-2/3) mũ 3

còn 81/125 chắc đề bài sai

-8/27=(-2/3)^3=-2^3/3^3

còn cái kia mk k bt

\(\frac{81}{125}\) viết dưới dạng luỹ thừa là : \(\frac{3^4}{5^3}\)

\(-\frac{8}{27}\) viết dưới dạng luỹ thừa là : \(-\frac{2^3}{3^3}\)

\(C=\dfrac{2^6\cdot3^{10}}{3^9\cdot2^6}=3\\ D=\dfrac{3^{24}\cdot3^{10}}{3^{21}\cdot3^{11}}=\dfrac{3^{34}}{3^{32}}=3^2=9\\ F=\dfrac{2^{45}\cdot5^{14}}{5^{15}\cdot2^{47}}=\dfrac{1}{2^2\cdot5}=\dfrac{1}{20}\\ G=\dfrac{2^2\cdot5^2\cdot5^3}{2^3\cdot5^4}=\dfrac{1\cdot5}{2}=\dfrac{5}{2}\)

C=3

D=9

F=1/20

G=5/2

Em ko giải chi tiết vì nó lâu

Mong thông cảm!

C= \(\dfrac{8^2.9^5}{3^9.4^3}\)

   = \(\dfrac{\left(2^3\right)^2.\left(3^2\right)^5}{3^9.\left(2^2\right)^3}\)

   = \(\dfrac{2^6.3^{10}}{3^9.2^6}\)

   = \(\dfrac{3^1}{1}\)

   = \(\dfrac{3}{1}\)

\(log_5125=log_55^3=3\)

\(log_6216=log_66^3=3\)

\(log_{10}\dfrac{1}{10000}=log_{10}10^{-4}=-4\)

\(log\sqrt{1000}=log_{10}10^{\dfrac{3}{2}}=\dfrac{3}{2}\)

\(81^{log_35}=3^{3log_35}=3^{log_3125}=125\)

\(125^{log_52}=5^{3log_52}=5^{log_58}=8\)

\(\left(\dfrac{1}{49}\right)^{log_7\dfrac{1}{8}}=7^{-2log_7\dfrac{1}{8}}=7^{log_764}=64\)

\(\left(\dfrac{1}{625}\right)^{log_52}=5^{-4log_52}=5^{log_5\dfrac{1}{16}}=\dfrac{1}{16}\)