計算: (1)(a-2b+c)(a+2b-c)-(a+2b+c)2; (2)(x+y)4(x-y)4; (3)(a+b+c)(a2+b2+c2-ab-ac-bc).
計算:
(1)(a-2b+c)(a+2b-c)-(a+2b+c)2;
(2)(x+y)4(x-y)4;
(3)(a+b+c)(a2+b2+c2-ab-ac-bc).
(1)(a-2b+c)(a+2b-c)-(a+2b+c)2;
(2)(x+y)4(x-y)4;
(3)(a+b+c)(a2+b2+c2-ab-ac-bc).
其他人氣:303 ℃時間:2020-06-17 21:41:49
優(yōu)質(zhì)解答
(1)原式=a2-(2b-c)2-(a2+4b2+c2+4ab+4bc+2ac)=a2-4b2-c2+4bc-a2-4b2-c2-4ab-4bc-2ac=-8b2-2c2-4ab-2ac;(2)原式=[(x+y)(x-y)]4=(x2-y2)4=x8+y8+6x4y4-4x6y2-4x2y6;(3)原式=a3+ab2+ac2-a2b-a2c-abc+...
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