由条件:2(a+2ab+b)=9990
4ab+2a+2b+1=9991
(2a+1)(2b+1)=9991=103*97=9991*1
2a+1=103 a=51
2b+1=97 b=48
2a+1=97 a=48
2b+1=103 b=51
2a+1=9991 a=4995
2b+1=1 b=0
2a+1=1 a=0
2b+1=9991 b=4995
则有序数对(a,b)有4对
由条件:2(a+2ab+b)=9990
4ab+2a+2b+1=9991
(2a+1)(2b+1)=9991=103*97=9991*1
2a+1=103 a=51
2b+1=97 b=48
2a+1=97 a=48
2b+1=103 b=51
2a+1=9991 a=4995
2b+1=1 b=0
2a+1=1 a=0
2b+1=9991 b=4995
则有序数对(a,b)有4对