Question

已知(a-2)²+(b+1)²=0求3a²b+ab²-3a²b+5ab+ab²-4ab+½a²b的值

Answer 1

(a-2)²+(b+1)²=0
a-2=0,b+1=0
a=2, b=-1
3a²b+ab²-3a²b+5ab+ab²-4ab+½a²b
=(3-3+½)a²b+(1+1)ab²+(5-4)ab
=½a²b+2ab²+ab
=ab(½a+2b+1)
=2×(-1)×﹙½×2-2+1)
=0

追问

(a-2)²+(b+1)²=0
a-2=0,b+1=0
为什么

追答

几个非负数的和等于0
只有当他们都是0时成立

Answer 2

(a-2)²+(b+1)²=0

所以a-2=b+1=0
a=2,b=-1

所以原式=2ab²+ab+½a²b
=4-2-2
=0

Answer 3

根据题意a-2=0 a=2 b+1=0 b=-1
3a²b+ab²-3a²b+5ab+ab²-4ab+½a²b=1/2a²b+2ab²+ab
=ab(1/2a+2b+1)=-2×(1-2+1)=0

追问

(a-2)²+(b+1)²=0
a-2=0,b+1=0
为什么根据那条定律

Answer 4

1. 3a²b+ab²-(3a²b-5ab)+ab²-4ab+0.5a²b
   =3a²b+ab²-3a²b+5ab+ab²-4ab+0.5a²b
   =0.5a²b+2ab²+ab
   =ab(0.5a+2b+1)
   带入a .b  就可

3a²b+ab²-(3a²b-5ab)+ab²-4ab+0.5a²b
=3a²b+ab²-3a²b+5ab+ab²-4ab+0.5a²b
=0.5a²b+2ab²+ab
=ab(0.5a+2b+1)
带入a .b 就可