题目内容
计算:
(1)y•yn+1-2yn•y2;
(2)5•(-5)2m+(-5)2m+1;
(3)(a-b)3(b-a)4.
(1)y•yn+1-2yn•y2;
(2)5•(-5)2m+(-5)2m+1;
(3)(a-b)3(b-a)4.
考点:同底数幂的乘法
专题:
分析:(1)先算同底数幂的乘法,再合并同类项即可;
(2)先变形为-(-5)•(-5)2m+(-5)2m+1,再算同底数幂的乘法,再合并同类项即可;
(3)先变形为(a-b)3(a-b)4,再算同底数幂的乘法.
(2)先变形为-(-5)•(-5)2m+(-5)2m+1,再算同底数幂的乘法,再合并同类项即可;
(3)先变形为(a-b)3(a-b)4,再算同底数幂的乘法.
解答:解:(1)y•yn+1-2yn•y2
=yn+2-2yn+2
=-yn+2;
(2)5•(-5)2m+(-5)2m+1
=-(-5)•(-5)2m+(-5)2m+1
=-(-5)2m+1+(-5)2m+1
=0;
(3)(a-b)3(b-a)4
=(a-b)3(a-b)4
=(a-b)7.
=yn+2-2yn+2
=-yn+2;
(2)5•(-5)2m+(-5)2m+1
=-(-5)•(-5)2m+(-5)2m+1
=-(-5)2m+1+(-5)2m+1
=0;
(3)(a-b)3(b-a)4
=(a-b)3(a-b)4
=(a-b)7.
点评:考查了同底数幂的乘法法则:同底数幂相乘,底数不变,指数相加.
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