急!!求学霸帮忙!!初一下数学题!!
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(1)解:原式=(5^m)^3*(5^n)^2
因为5^m=3
5^n=2
所以原式=3……3*2*4
=27*4
=108
所以5^3m+2n=108
5^2m+n+5
=(5^m)^2*5^n*5
=3^2*2*5
=9*2*5
=90
(2)解原式=2^100+2^100
=2*(1+100)
=2^101
原式=2^100-2^101
=2^100*(1-2)
=-2^100
(3)解原式=(a-b)^2n+1*(a-b)^2n
=(a-b)^(2n+1+2n)
=(a-b)^4n+1
(4)解:因为x*(3+n)=x^(1+2n+1)
x*(3+n)=x^(2n+2)
所以3+n=2n+2
n=1
(5)解原式=x^(1+4+3)
=x^8
解原式=-x^(2+2+4)
=-x^8
解原式=b^(2+2+3)
=b^7
因为5^m=3
5^n=2
所以原式=3……3*2*4
=27*4
=108
所以5^3m+2n=108
5^2m+n+5
=(5^m)^2*5^n*5
=3^2*2*5
=9*2*5
=90
(2)解原式=2^100+2^100
=2*(1+100)
=2^101
原式=2^100-2^101
=2^100*(1-2)
=-2^100
(3)解原式=(a-b)^2n+1*(a-b)^2n
=(a-b)^(2n+1+2n)
=(a-b)^4n+1
(4)解:因为x*(3+n)=x^(1+2n+1)
x*(3+n)=x^(2n+2)
所以3+n=2n+2
n=1
(5)解原式=x^(1+4+3)
=x^8
解原式=-x^(2+2+4)
=-x^8
解原式=b^(2+2+3)
=b^7
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