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计算:
(1)2-2×(42×80
(2)(-m32•(-2m23÷(2m)2
(3)(-3a2)•(2a-1)+a(1-3a)
(4)(x-2y)(x2+2xy+4y2
人气:138 ℃ 时间:2020-03-24 12:09:11
解答
(1)2-2×(42×80)=
1
4
×16×1=4;
(2)(-m32•(-2m23÷(2m)2=m6•(-8m6)÷4m2=-2m10
(3)(-3a2)•(2a-1)+a(1-3a)=-6a3+3a2+a-3a2=-6a3+a;
(4)(x-2y)(x2+2xy+4y2)=x3+2x2y+4xy2-2x2y-4xy2-8y3=x3-8y3
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