(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
(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