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计算:
(1)(-
1
2
)2÷(-2)-3+2-2×(-3)0

(2)(-b)7÷b3•(-b)2÷(-b23
(3)(3x2y-2x+1)(-2xy)
(4)(x-3y)(y+3x)-(x-3y)(3y-x)
人气:181 ℃ 时间:2020-02-05 23:51:52
解答
(1)原式=
1
4
÷(-
1
8
)+
1
4
×1=-2+
1
4
=-
7
4

(2)原式=-b7÷b3•b2÷(-b6)=-b4•b2÷(-b6)=-b6÷(-b6)=1;
(3)原式=-6x3y2+4x2y-2xy;
(4)原式=xy+3x2-3y2-9xy+x2-6xy+9y2=4x2-14xy+6y2
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