解：∵(x+y)(x+y-1)-12=0,∴(x+y)²-(x+y)-12=0[(x+y)-4][(x+y)+3]=0(x+y)-4=0 或 (x+y)+3=0∴x+y=4 或 x+y=-3
令x+y=t则t(t-1)-12=0t²-t-12=0(t-4)(t+3)=0t1=4,t2=-3即x+y=4或x+y=-3
