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TT
0
AH
20 tháng 11 2016
Câu 1 :
\(\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^5}\)
= \(\frac{5^{32}.2^{86}.5^{43}}{\left(-2\right)^{87}.5^{15}}\)
= \(\frac{5^{72}.\left(-2\right)^{86}}{\left(-2\right)^{87}.5^{75}}\)
= \(\frac{1}{-2}\)
Câu 2 :
\(\frac{5^4.18^4}{125.9^5.16}\)
= \(\frac{5^4.2^4.3^8}{5^3.3^{10}.2^4}\)
= \(\frac{5}{3^2}\)
= \(\frac{5}{9}\)
Câu 3 :
\(\frac{9^{18}.2^{29}}{8^9.27^{12}}\)
= \(\frac{3^{36}.2^{29}}{2^{27}.3^{36}}\)
= \(2^2\)
= 4
H
17 tháng 8 2021
a) 9/18 - (-7/12) + 13/32
= 13/12 + 13/32
= 143/96
b) (5/-8) + 14/25 - 6/10
= (-13/200) - 6/10
= -133/200
Chúc bạn học tốt!! ^^
17 tháng 8 2021
a: \(\dfrac{9}{18}-\dfrac{-7}{12}+\dfrac{13}{32}\)
\(=\dfrac{1}{2}+\dfrac{7}{12}+\dfrac{13}{32}\)
\(=\dfrac{48}{96}+\dfrac{56}{96}+\dfrac{39}{96}\)
\(=\dfrac{143}{96}\)
b: \(\dfrac{-5}{8}+\dfrac{14}{25}-\dfrac{6}{10}\)
\(=\dfrac{-125}{200}+\dfrac{112}{200}-\dfrac{120}{200}\)
\(=\dfrac{-133}{200}\)
27 tháng 8 2017
a)\(\frac{5}{3}\)+ \(\left(\frac{-2}{7}\right)\)-(-1,2)
=\(\frac{5}{3}+\left(\frac{-2}{7}\right)+\frac{6}{5}\)
=\(\frac{175+\left(-30\right)+126}{105}\)
=\(\frac{271}{105}\)
b) \(\frac{-4}{9}+\frac{-5}{6}-\frac{17}{4}\)
=\(\frac{-16+\left(-30\right)-153}{36}\)
=\(\frac{-199}{36}\)
14 tháng 12 2017
\(\frac{5}{3}+\left(\frac{-2}{7}\right)-\left(\frac{-6}{5}\right)\)
=\(\frac{-2}{7}-\left(\frac{5}{3}+\frac{-6}{5}\right)\)
=\(\frac{-79}{105}\)\(\frac{-2}{7}-\frac{7}{15}\)
27 tháng 7 2022
\(=5+\dfrac{1}{5}-\dfrac{2}{9}-2+\dfrac{1}{23}+\dfrac{73}{35}-\dfrac{5}{6}-8-\dfrac{2}{7}+\dfrac{1}{18}\)
\(=-5+\left(\dfrac{1}{5}-\dfrac{2}{7}+\dfrac{73}{35}\right)+\left(\dfrac{-2}{9}-\dfrac{5}{6}+\dfrac{1}{18}\right)+\dfrac{1}{23}\)
\(=-5+\dfrac{7-10+73}{35}+\dfrac{-4-15+1}{18}+\dfrac{1}{23}\)
\(=-5+2-1+\dfrac{1}{23}=-4+\dfrac{1}{23}=-\dfrac{91}{23}\)
15 tháng 12 2017
a) \(\frac{15}{12}+\frac{5}{13}-\frac{3}{12}-\frac{18}{13}\)
\(=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)
\(=1+\left(-1\right)\)
\(=0\)
b) \(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(20.5\right)^4}{\left(25.4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)
c) \(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{12}.\left(2^{18}+2^8\right)}{2^{12}.\left(1+2^{10}\right)}=\frac{2^{18}+2^8}{1+2^{10}}=256\)
`#3107.101107`
\(\dfrac{2^8\cdot2^{18}}{8^5\cdot4^6}=\dfrac{2^{8+18}}{\left(2^3\right)^5\cdot\left(2^2\right)^6}=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{2^{26}}{2^{15+12}}=\dfrac{2^{26}}{2^{27}}=\dfrac{1}{2}\)
\(\dfrac{2^8.2^{18}}{8^5.4^6}=\dfrac{2^{8+18}}{\left(2^3\right)^5.\left(2^2\right)^6}\\ =\dfrac{2^{26}}{2^{15}.2^{12}}=\dfrac{2^{26}}{2^{15+12}}\\ =\dfrac{2^{26}}{2^{27}}=\dfrac{1}{2^1}=\dfrac{1}{2}\)