\(\dfrac{27^4.4^3}{9^5.8^2}\);\(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}\)
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HK
2
25 tháng 12 2021
\(1.\dfrac{27^4.4^3}{9^5.8^2}=\dfrac{3^{12}.2^6}{3^{10}.2^6}=3^2=9\)
\(2.\dfrac{8^5.3^{15}}{2^{14}.81^4}=\dfrac{2^{15}.3^{15}}{2^{14}.3^{16}}=\dfrac{2}{3}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
25 tháng 12 2021
Ý 1:
\(\dfrac{27^4.4^3}{9^5.8^2}=\dfrac{\left(3^3\right)^4.\left(2^2\right)^3}{\left(3^2\right)^5.\left(2^3\right)^2}=\dfrac{3^{12}.2^6}{3^{10}.2^6}=3^2=9\)
Ý 2:
\(\dfrac{8^5.3^{15}}{2^{14}.81^4}=\dfrac{\left(2^3\right)^5.3^{15}}{2^{14}.\left(3^4\right)^4}=\dfrac{2^{15}.3^{15}}{2^{14}.3^{16}}=\dfrac{2^{14}.2.3^{15}}{2^{14}.3^{15}.3}=\dfrac{2}{3}\)
30 tháng 8 2022
a: \(=\dfrac{2^5\cdot2^{12}\cdot2^6}{2^{24}}=\dfrac{1}{2}\)
b: \(=\dfrac{12-15}{20}\cdot\left(\dfrac{10-6}{30}\right)^2\)
\(=\dfrac{-3}{20}\cdot\left(\dfrac{2}{15}\right)^2=\dfrac{-3}{20}\cdot\dfrac{4}{225}=-\dfrac{1}{375}\)
c: \(=3\cdot\dfrac{2}{5}+2:\dfrac{1}{4}=1.2+8=9.2\)
10 tháng 8 2023
\(=\dfrac{3^8\cdot2^{10}\cdot5^6}{2^9\cdot3^6\cdot5^7}=3^2\cdot\dfrac{2}{5}=\dfrac{18}{5}\)
10 tháng 8 2023
\(\dfrac{9^4\cdot4^5\cdot25^3}{8^3\cdot27^2\cdot5^7}\)
\(=\dfrac{\left(3^2\right)^4\cdot\left(2^2\right)^5\cdot\left(5^2\right)^3}{\left(2^3\right)^3\cdot\left(3^3\right)^2\cdot5^7}\)
\(=\dfrac{3^8\cdot2^{10}\cdot5^6}{2^9\cdot3^6\cdot5^7}\)
\(=\dfrac{3^2\cdot2\cdot1}{1\cdot1\cdot5}\)
\(=\dfrac{18}{5}\)
20 tháng 12 2018
\(\frac{27^4.4^3}{9^5.8^2}=\frac{\left(3^3\right)^4.\left(2^2\right)^3}{\left(3^2\right)^5.\left(2^3\right)^2}=\frac{3^{12}.2^6}{3^{10}.2^6}=3^2=9\)
20 tháng 12 2018
=\(\frac{3^{12}.2^6}{3^{15}.2^6}\)
=\(\frac{3^{12}}{3^{12}.3^3}\)
=\(\frac{1}{27}\)
NN
0
1 tháng 4 2023
\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)
= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\)
\(=\dfrac{1}{2}-\dfrac{1}{17}\)
\(=\dfrac{15}{34}\)
Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)
a)\(\dfrac{27^4.4^3}{9^5.8^2}\)
=\(\dfrac{3^{12}.2^6}{3^{10}.2^6}\)
=3\(^2\)=9
b)\(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}\)
=\(\dfrac{3^{29}.2^{32}}{3^{27}.2^{33}}\)
=\(\dfrac{9}{2}\)
\(\dfrac{27^4.4^3}{9^5.8^2}=\dfrac{\left(3^3\right)^4.\left(2^2\right)^3}{\left(3^2\right)^5.\left(2^3\right)^2}=\dfrac{3^{12}.2^6}{3^{10}.2^6}=\dfrac{3^{12}}{3^{10}}=3^2=9\)
_________
\(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}=\dfrac{3^{29}.\left(2^2\right)^{16}}{\left(3^3\right)^9.\left(2^3\right)^{11}}=\dfrac{3^{29}.2^{32}}{3^{27}.2^{33}}=\dfrac{1}{3^2.2}=\dfrac{1}{9.2}=\dfrac{1}{18}\)