a^2=1.a^m=2,a^n=3,a^p=4,求a^m+n+p+4的值,补充一道:若5^2x+1=125,求(x-2)^2014+x的值
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a^2=1.a^m=2,a^n=3,a^p=4,求a^m+n+p+4的值,补充一道:若5^2x+1=125,求(x-2)^2014+x的值
a^2=1.a^m=2,a^n=3,a^p=4,求a^m+n+p+4的值,
补充一道:若5^2x+1=125,求(x-2)^2014+x的值
a^2=1.a^m=2,a^n=3,a^p=4,求a^m+n+p+4的值,补充一道:若5^2x+1=125,求(x-2)^2014+x的值
答:
a^2=1.a^m=2,a^n=3,a^p=4,
所以:a^4=(a^2)^2=1
所以:
a^(m+n+p+4)
=(a^m)*(a^n)*(a^p)*(a^4)
=2*3*4*1
=24
5^2x+1=125,
5^(2x+1)=5^3
2x+1=3
2x=2
x=1
(x-2)^2014+x
=(1-2)^2014+1
=1+1
=2
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