Question

已知a^2+4ab+b^2+a^2b^2+1=0，求a^3b+2a^2b^2+ab^3-ab的值

帮帮忙 解一下

Answer 1

(a²b²+2ab+1)+(a²+2ab+b²)=0
(ab+1)²+(a+b)²=0
所以ab+1=a+b=0
ab=-1,a+b=0
所以a=1,b=-1或a=-1,b=1

原式ab*a²+2(ab)²+ab*b²-ab
=-1+2-1+1
=1

Answer 2

a^2+4ab+b^2+a^2b^2+1=0
a^2+2ab+b^2+a^2b^2+2ab+1=0
(a+b)^2+(ab+1)^2=0
a+b=0
ab=-1

a^3b+2a^2b^2+ab^3-ab
=ab(a^2+2ab+b^2)-ab
=ab(a+b)^2-ab
=0+1
=1

Answer 3

a^2+4ab+b^2+a^2b^2+1=0
(a+b)^2+(ab+1)^2=0
所以 a=-b ab=-1
a^3b+2a^2b^2+ab^3-ab
ab(a^2+2ab+b^2-1)
=ab[(a+b)^2-1]
=-1x(0-1)
=1

Answer 4

已知因式分解得（a+b)²+（ab+1）²=0，所以a+b=0,ab=-1
求解式=ab（a²+2ab+b²-1）=ab（a+b）²-ab=1