∵f（x）是二次函数，∴设f（x）=ax2+bx+c，则f（x+2）=a（x+2）2+b（x+2）+c=ax2+（4a+b）x+4a+2b+c，∴f（x+2）+f（x）=2ax2+（4a+2b）x+4a+2b+2c，∵f（x+2）+f（x）=2x2+5x+5，∴2a=24a+2b=54a+2b+2c=5，解得a=1...