∵方程有实数根∴判别式大于等于0△=[2(1+a)]²-4(3a²+4ab+4b²+2)=[4(1+2a+a²)]-4(3a²+4ab+4b²+2)=(4+8a+4a²)-(12a²+16ab+16b²+8)=4+8a+4a²-12a²-16ab-16b&#...