1，F(x)是二次函数，且f(x+1)+f(x-1)=2x平方-4x求f (1-根号2)
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1:设y=ax^2+bx+c,则a(x+1)^2+b(x+1)+c+a(x-1)^2+b(x-1)+c=2x^2-4x即2ax^2+2bx+2a^2+2c=2x^2-4x,比较系数得:2a=2,2b=-4,2a^2+2c=0,所以a=1,b=-2,c=-1,所以y=x^2-2x-1,所以f(1-√2)=02:f(-7.6)=f(-7.6+4)=f(-3.6)=f(-3.6+4)=f(0.4)=0.4