已知f(x)为二次函数,且f(x+1)+f(x-1)=2x²-4x,求f(1-√2)

f(x+1)+f(x-1)=2x²-4x
f(x)=ax^2+bx+c f(x+1)=a(x+1)^2+b(x+1)+c=ax^2+(2a+b)x+a+b+c
f(x-1)=a(x-1)^2+b(x-1)+c=ax^2+(b-2a)x+a-b+c f(x+1)+f(x-1)=ax^2+(2a+b)x+a+b+c+ax^2+(b-2a)x+a-b+c =2ax^2+2bx+2a+2c=2x^2-4x 2a=2 a=1 2b=-4 b=-2 2a+2c=0 c=-1
f(x)=x^2-2x-1=x^2-2x+1-2=(x-1)^2-2 f(1-√2)=(1-√2-1)^2-2=2-2=0

设f(x)=ax²+bx+c
∴f(x+1）+f(x-1)
=a(x+1)²+b(x+1)+c+a(x-1)²+b(x-1)+c
=2ax²+2bx+2a+2c
=2x²-4x
∴2a=2,2b=-4,2a+2c=0
a=1,b=-2,c=-1
∴f(x)=x²-2x-1
f(1-√2)=(1-√2)²-2(1-√2)-1=1+2-2√2-2+2√2-1=0
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