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初一数学计算(要有过程)
(1+2ab)²-(2+2ab)²=
(7x+y)(7x-y)-(-x+y)(x-y)=
(nm+2分之1)(nm-2分之1)²=
(x-2y-1)(x+2y-1)=
(a-b-2c)²=
人气:498 ℃ 时间:2020-06-04 15:46:56
解答
(1+2ab)²-(2+2ab)²=[(1+2ab)+(2+2ab)]*[(1+2ab)-(2+2ab)]=(1+2ab+2+2ab)*(1+2ab-2-2ab)=(3+4ab)*(-1)=-3-4ab(7x+y)(7x-y)-(-x+y)(x-y)=(7x)²-y²+(x-y)(x-y)=(7x)²-y...(nm+2分之1)(nm-2分之1)²=(x-2y-1)(x+2y-1)=(a-b-2c)²=(x-2y-1)(x+2y-1)=(x-1-2y)(x-1+2y)=[(x-1)-2y][(x-1)+2y]=(x-1)²-4y²=x²-2x+1-4y² (a-b-2c)²=[(a-b)-2c]²=(a-b)²-4c(a-b)+4c²=a²-2ab+b²-4ac+4ab+c²=a²+b²+c²-2ab-4ac+4ab (nm+2分之1)(nm-2分之1)²=不会
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