计算. (1)(a-2b+3c)2-(a+2b-3c)2; (2)[ab(3−b)−2a(b−1/2b2)](−3a2b3); (3)-2100×0.5100×(-1)2013÷(-1)-5; (4)[(x+2y)(x-2y)+4(x-y)
问题描述:
计算.
(1)(a-2b+3c)2-(a+2b-3c)2;
(2)[ab(3−b)−2a(b−
b2)](−3a2b3);1 2
(3)-2100×0.5100×(-1)2013÷(-1)-5;
(4)[(x+2y)(x-2y)+4(x-y)2-6x]÷6x;
(5)5a2-[a2+(5a2-2a)-2(a2-3a)].
答
(1)(a-2b+3c)2-(a+2b-3c)2=(a-2b+3c+a+2b-3c)(a-2b+3c-a-2b+3c)=2a•(-4b+6c)=12ac-8ab;
(2)[ab(3−b)−2a(b−
b2)](−3a2b3)=[3ab-ab2-2ab+ab2](-3a2b3)=ab(-3a2b3)=-3a3b4;1 2
(3)-2100×0.5100×(-1)2013÷(-1)-5=-(2×0.5)100×(-1)÷(-1)=-1×(-1)÷(-1)=-1;
(4)[(x+2y)(x-2y)+4(x-y)2-6x]÷6x=[(x2-4y2)+4(x2-2xy+y2)-6x]÷6x=[x2-4y2+4x2-8xy+4y2-6x]÷6x=[5x2-8xy-6x]÷6x=
x-5 6
y-1;4 3
(5)5a2-[a2+(5a2-2a)-2(a2-3a)]=5a2-[a2+5a2-2a-2a2+6a]=5a2-[4a2+4a]=a2-4a.