1.viết gọn các tích sau dưới dạng lũy thừa. a.3.5.5.3.3.5. b.2.2.10.10.5 C.3.5.27.125
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a, 2.2.10.10.5 = 2.(2.5).10.10 = 2.10.10.10 = 2.10^3
b, 3.5.27.125 = 3.5.3^3.5^3 = 3^4.5^4
c, 2.3.12.12.3 = 2.3.3.2^2.3.2^2.3 = 2^5.3^4
d, 2.3.8.12.24 = 2.3.2^3.3.2^2.2^3.3 = 2^9.3^3
A)2.2.10.10.5=2.(2.5).10.10
=2.10.10.10=21.103
B) 3.5.27.125=3.5.(9.3).(25.5)
=3.5.3.3.3.5.5.5=3.3.3.3.5.5.5.5=34.54
C) 2.3.12.12.3=2.3.(4.3).(4.3).3
=2.3.2.2.3.2.2.3.3=2.2.2.2.2.3.3.3.3=25.34
Đ)2.3.8.12.24=2.3.(2.2).(3.4).(6.2)
=2.3.2.2.3.2.2.3.2.2=27.33
a) \(a^6\)
b) 4.5.20.20.20=20.20.20.20=\(20^4\)
c) 3.2.12.12.3=6.6.6.6.6.3=\(6^5\).3= \(6^5\),\(6^1\)=\(6^6\)
Phần d mik ko bt lm
a) \(2.2.2.2.3.3.3.3=2^4.3^4\)
b) \(2.33.5555=2^1.3^2.5^4\)
c) \(5.5.5.5.4.4.4=5^4.4^3\)
d) \(8.8.10.25.16=2^3.2^3.2.5.5^2.2^4=2^{11}.5^3\)
a: \(3\cdot3\cdot3\cdot3\cdot3=3^5\)
b: \(y\cdot y\cdot y\cdot y=y^4\)
c: \(5\cdot p\cdot5\cdot p\cdot2\cdot q\cdot4\cdot q=25\cdot2\cdot4\cdot p^2q^2=2\cdot\left(10qp\right)^2\)
d: \(a\cdot a+b\cdot b+c\cdot c+d\cdot d\cdot d\cdot d=a^2+b^2+c^2+d^4\)
a) 3.5.5.3.3.5 = 33.53
b) 2.2.10.10.5 = 2.2.2.5.2.5.5 = 24.53
c) 3.5.27.125 = 3.5.33.53 = 34.54
a,3.5.5.3.3.5=33.53
b,2.2.10.10.5=22.102.5
c,3.5.27.125=3.5.3.3.3.5.5.5=34.54