令f(x)=ax^2 +bx +c;
则f(x+1)+f(x+2)=2ax^2 +(6a+2b)x +(5a+3b+2c);
即:2ax^2 +(6a+2b)x +(5a+3b+2c)=2x^2-4x
分别对应各项的系数,得
2a=2;
6a+2b=-4;
5a+3b+2c=0;
解得:
a=1; b=-5; c=5;
则 f(x)=x^2 -5x +5.
则f(1-根号2）=(3-2√2) -5+5√2 +5
=3+3√2