已知(a+b)²+ 2b-1的绝对值=0 求多项式ab-[2ab-3(ab-1)]的值


速度啊!初一的数学题!
(a+b)²+ 2b-1的绝对值=0
满足 a+b = 0 2b-1 = 0
b = 1/2 a = -1/2

求多项式ab-[2ab-3(ab-1)]的值
= ab - 2ab + 3ab - 3
= 2ab -3
= 2 x 1/2 x (-1/2) -3
= -1/2 -3
= - 2分之 7

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(薯脊a+b)²春早+ 2b-1的绝对值=0
所以a+b=0,2b-1=0
所以b=1/2
a=-b=-1/2
所以原式=ab-2ab+3ab-3
=2ab-3
=-1/2-3
=-7/扒手雀2
(a+b)²+ 2b-1的好咐绝对值=0
则:a+b=0,2b-1=0
解得:b=1/2,a=-1/2
那么悉笑,ab-[2ab-3(ab-1)]=ab-2ab+3ab-3=2ab-3=2×(-1/2)×1/2-3=-3又友陆纯1/2
(a+b)^2=0
2b-1=0
所耐消以a= -0.5 b=0.5
所以多项式ab-【历亩丛2ab-3(ab-1)肢樱】= -3.5
a=1,b=0,结果为-3