∵f(x+2)=f(2-x)∴对称轴为x=2设二次函数为f(x)=a(x-2)²+c=ax²-4ax+4a+c (a≠0)∴当△=16a²-16a²-4ac=-4ac≥0时,f(x)=0存在两个实数根.设为x1,x2∴x1+x2=4,x1x2=4+c/a∴x1²+x2²=(x1+x...