f(x+y)+f(x-y)=2f(x)f(y) 求证：f(0)=1
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f(x+y)+f(x-y)=2f(x)f(y)求证：f(0)=1f(x+y)+f(x-y)=2f(x)f(y)求证：f(0)=1f(x+y)+f(x-y)=2f(x)f(y)求证：f(0)=1令x=y
f(x+y)+f(x-y)=2f(x)f(y) 求证：f(0)=1
f(x+y)+f(x-y)=2f(x)f(y) 求证：f(0)=1
f(x+y)+f(x-y)=2f(x)f(y) 求证：f(0)=1
令x=y=0
则x+y=x-y=0
所以f(0)+f(0)=2[f(0)]²
2f[f(0)]²-2f(0)=0
f(0)[f(0)-1]=0
f(0)=0,f(0)=1
如果题目里说f(0)≠0
则f(0)=1,否则就有两个解
f(1+0)+f(1-0)=2f(1)f(0)
得f(1)+f(1)=2f(1)f(0)
得2f(1)=2f(1)f(0)
得f(0)=1
设x=0,y=0则带入得f(0)+f(o)=2f(0)f(0)即2f(0)=2f(0)f(0)即f(0）=1或f(0)=0
设x=1,y=0则带入得f(1)+f(1)=2f(1)f(0)即2f(1)=2f(1)f(0),而f(1)不等于0
则f(0)=1
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