试题

题目:
x2m=2,xn=5,求x4m+2n的值.
答案
解:∵x2m=2,xn=5,
∴x4m+2n
=x4m·x2n
=(x2m2·(xn2
=22×52
=100.
解:∵x2m=2,xn=5,
∴x4m+2n
=x4m·x2n
=(x2m2·(xn2
=22×52
=100.
考点梳理
幂的乘方与积的乘方;同底数幂的乘法.
根据同底数幂的乘法得出x4m·x2n,根据幂的乘方化成(x2m2·(xn2,代入求出即可.
本题考查了同底数幂的乘法,幂的乘方的应用,用了整体代入思想.
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