题目内容
13、计算:
(1)(3a-2b)(9a+6b);
(2)(2y-1)(4y2+1)(2y+1)
(1)(3a-2b)(9a+6b);
(2)(2y-1)(4y2+1)(2y+1)
分析:根据平方差公式(a+b)(a-b)=a2-b2,即可解答本题.
解答:解:(1)(3a-2b)(9a+6b)
=3(3a+2b)(3a-2b)
=3[(3a)2-(2b)2]
=27a2-12b2;
(2)(2y-1)(4y2+1)(2y+1)
=(4y2-1)(4y2+1)
=16y4-1.
=3(3a+2b)(3a-2b)
=3[(3a)2-(2b)2]
=27a2-12b2;
(2)(2y-1)(4y2+1)(2y+1)
=(4y2-1)(4y2+1)
=16y4-1.
点评:本题考查了平方差公式的运用,比较简单.
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