解:原方程变形为:(x+y)(x-y+1)=12,
∵x,y为整数,
∴当x+y=3,x-y+1=4,解得⎧⎪⎪⎨⎪⎪⎩x=3y=0;
当x+y=4,x-y+1=3,解得⎧⎪⎪⎨⎪⎪⎩x=3y=1;
当x+y=-3,x-y+1=-4,解得⎧⎪⎪⎨⎪⎪⎩x=-4y=1;
当x+y=-4,x-y+1=-3,解得⎧⎪⎪⎨⎪⎪⎩x=-4y=0;
当x+y=2,x-y+1=6,解得⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩x= 72y=- 32(舍去) ;
当x+y=6,x-y+1=2,解得⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩x= 72y= 52(舍去);
当x+y=-2,x-y+1=-6,解得⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩x=- 92y= 52(舍去);
当x+y=-6,x-y+1=-2,解得⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩x=- 92y=- 32(舍去) .
当x+y=1,x-y+1=12,解得⎧⎪⎪⎨⎪⎪⎩x=6y=-5 ;
当x+y=12,x-y+1=1,解得⎧⎪⎪⎨⎪⎪⎩x=6y=6 ;
当x+y=-1,x-y+1=-12,解得⎧⎪⎪⎨⎪⎪⎩x=-7y=6 ;
当x+y=-12,x-y+1=-1,解得⎧⎪⎪⎨⎪⎪⎩x=-7y=-5 ;
故答案为:C
故答案为:c
选D.方程左侧因式分解(x+y)(x-y+1)=12的12个方程组x+y和x+y+1分别为(1,12)(-1,-12)(12,1)(-12,-1)(2,6)(-2,-6)(6,2)(-6,-2)(3,4)(-3,-4)(4,3)(-4,-3)共12个方程组
好