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2 tháng 10 2021
a: \(\left(0.5\right)^3\cdot2^3=1\)
b: \(\left(0.25\right)^2\cdot16=1\)
c: \(\left(\dfrac{3}{5}\right)^3:\left(-\dfrac{27}{1000}\right)=\dfrac{3^3}{5^3}\cdot\dfrac{-1000}{27}=\dfrac{-1000}{125}=-8\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 7 2023
a) \(\dfrac{27^3\cdot11+9^5\cdot5}{3^9\cdot2^4}\)
\(=\dfrac{3^9\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\)
\(=\dfrac{3^9\cdot\left(11+3\cdot5\right)}{3^9\cdot2^4}\)
\(=\dfrac{11+15}{16}\)
\(=\dfrac{26}{16}\)
\(=\dfrac{13}{8}\)
b) \(\dfrac{5^8+2^2\cdot25^4+2^3\cdot125^3-15^4\cdot5^4}{4^2\cdot625^2}\)
\(=\dfrac{5^8+2^2\cdot5^8+2^3\cdot5^9-3^4\cdot5^4\cdot5^4}{2^4\cdot5^8}\)
\(=\dfrac{5^8\cdot\left(1+2^2+2^3\cdot5-3^4\right)}{5^8\cdot2^4}\)
\(=\dfrac{1+4+40-81}{16}\)
\(=\dfrac{-36}{16}\)
\(=\dfrac{-9}{4}\)
c) \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)
\(=\dfrac{-2}{6}\)
\(=-\dfrac{1}{3}\)
4 tháng 10 2021
\(\dfrac{4^5.21+4^5.19}{2^{10}.11+2^{10}.5}=\dfrac{4^5.\left(21+19\right)}{2^{10}.\left(11+5\right)}=\dfrac{4^5.30}{2^{10}.16}\)\(=\dfrac{\left(2.2\right)^5.15}{2^{10}.8}=\dfrac{2^5.2^5.15}{2^{10}.8}=\dfrac{15}{8}\)
4 tháng 10 2021
\(\dfrac{4^2.25^2+32.125}{2^3.5^2}=\dfrac{2^4.5^4+2^5.5^3}{2^3.5^2}=\dfrac{2^3.5^2\left(2.5^2+2^2.5\right)}{2^3.5^2}=2.5^2+2^2.5=50+20=70\)
27 tháng 5 2022
b:\(=\dfrac{3^3\cdot2^3+3\cdot2^2\cdot3^2+3^3}{-13}=\dfrac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)
a: \(=\dfrac{5^5\cdot4^5\cdot5^{10}}{10^{20}}=\dfrac{5^{15}\cdot2^{10}}{5^{20}\cdot2^{20}}=\dfrac{1}{5^5\cdot2^{10}}\)
TC
27 tháng 5 2022
b) \(\dfrac{6^3+3.6^2+3^3}{-13}=\dfrac{2^3.3^3+3.3^2.2^2+3^3}{-13}=\dfrac{3^3\left(2^3+2^2+1\right)}{-13}\)
\(=\dfrac{3^3.\left(8+4+1\right)}{-13}=\dfrac{3^3.13}{-13}=\dfrac{27}{-1}=-27\)
a)\(\dfrac{20^5.5^{10}}{100^5}=\dfrac{20^5.5^{10}}{20^5.5^5}=\dfrac{5^5}{1}=3125\)
Y
15 tháng 8 2023
\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{2}\)
\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)
Vậy \(B=3\)
FC
18 tháng 10 2018
a, (5-5)-1 . \(\left(\dfrac{1}{2}\right)^{-2}\) . \(\dfrac{1}{10^5}\)
= 55 . \(\dfrac{1^{-2}}{2^{-2}}\) . \(\dfrac{1}{10^5}\)
= (55 . \(\dfrac{1}{10^5}\)) . \(\dfrac{1}{4}\)
= \(\dfrac{5^5}{10^5}\) . \(\dfrac{1}{4}\) = \(\left(\dfrac{1}{2}\right)^5\). \(\dfrac{1}{4}\)
= \(\dfrac{1}{32}.\text{}\dfrac{1}{4}\)= \(\dfrac{1}{128}\)
21 tháng 10 2022
b: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\dfrac{4}{5}\)
c: \(=\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^3\cdot5^2}=\dfrac{2^4\cdot5^3\left(5+2\right)}{2^3\cdot5^2}=10\cdot7=70\)
17 tháng 9 2023
\(a.\left[-\dfrac{6}{11}.\dfrac{11}{-6}\right].\dfrac{7}{10}.\left(-20\right)=1.7.\left(-2\right)=-14\)
\(b.\dfrac{-1}{2}:\dfrac{3}{4}.\dfrac{-7}{2}=\dfrac{7}{4}:\dfrac{3}{4}=\dfrac{7}{3}\)
\(c.\dfrac{93}{7}:-\dfrac{8}{9}+\dfrac{19}{7}:\dfrac{-8}{9}=\left(\dfrac{93}{7}+\dfrac{19}{7}\right):-\dfrac{8}{9}=\dfrac{-9}{8}.\dfrac{112}{7}=-18\)
\(=\dfrac{2^4\cdot5^4+2^5\cdot5^3}{2^{10}\cdot16}=\dfrac{2^4\cdot5^3\left(5+2\right)}{2^{10}\cdot2^4}=\dfrac{2^4\cdot5^3\cdot7}{2^{14}}=\dfrac{5^3\cdot7}{2^{10}}=\dfrac{875}{1024}\)