题目内容
计算:
(1)[a+(b-c)]•[a-(b-c)];
(2)(a-2b+3c)(a+2b-3c).
(1)[a+(b-c)]•[a-(b-c)];
(2)(a-2b+3c)(a+2b-3c).
(1)原式=a2-(b-c)2
=a2-(b2-2bc+c2)
=a2-b2+2bc-c2.
(2)原式=[a-(2b-3c)][a+(2b-3c)]
=a2-(2b-3c)2
=a2-4b2+12bc-9c2.
=a2-(b2-2bc+c2)
=a2-b2+2bc-c2.
(2)原式=[a-(2b-3c)][a+(2b-3c)]
=a2-(2b-3c)2
=a2-4b2+12bc-9c2.
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