Question

初一因式分解。若a^2+b^2+a^2b^2+4ab+1=0求a-b

Answer 1

a^2+b^2+a^2b^2+4ab+1=0
(a²+b²+2ab)+(a²b²+2ab+1)=0
(a+b)²+(ab+1)²=0

a+b=0且ab+1=0
则,a+b=0, ab=-1

(a-b)²
=(a+b)²-4ab
=0+4
=4

则,a-b=±2

Answer 2

a^2+b^2+a^2b^2+4ab+1=0
a²+b²+2ab+a²b²+2ab+1=0
(a+b)²+(ab+1)²=0
所以a+b=0，ab+1=0
解得a=1,b=-1或a=-1,b=1
所以a-b=1-（-1）=2或a-b=-1-1=-2

Answer 3

a^2+b^2+a^2b^2+4ab+1=0
(a^2+2ab+b^2)+(a^2b^2+2ab+1)=0
(a+b)^2+(ab+1)=0
a+b=0 且ab+1=0,
∴a=1，b=-1，或a=-1，b=1。