已知函数f(x)满足f(x+y)+f(x-y)=2f(x)(y) (x.y∈R)且f(0)≠0不等于零,证f(x)是偶函数.注:有改动(抱歉)已知函数f(x)满足f(x+y)+f(x-y)=2f(x)*f(y) (x.y∈R)且f(0)≠0不等于零,证f(x)是偶函数.
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![已知函数f(x)满足f(x+y)+f(x-y)=2f(x)(y) (x.y∈R)且f(0)≠0不等于零,证f(x)是偶函数.注:有改动(抱歉)已知函数f(x)满足f(x+y)+f(x-y)=2f(x)*f(y) (x.y∈R)且f(0)≠0不等于零,证f(x)是偶函数.](/uploads/image/z/5343828-60-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3f%28x%2By%29%2Bf%28x-y%29%3D2f%28x%29%28y%29+%28x.y%E2%88%88R%29%E4%B8%94f%280%29%E2%89%A00%E4%B8%8D%E7%AD%89%E4%BA%8E%E9%9B%B6%2C%E8%AF%81f%28x%29%E6%98%AF%E5%81%B6%E5%87%BD%E6%95%B0.%E6%B3%A8%EF%BC%9A%E6%9C%89%E6%94%B9%E5%8A%A8%EF%BC%88%E6%8A%B1%E6%AD%89%EF%BC%89%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3f%28x%2By%29%2Bf%28x-y%29%3D2f%28x%29%2Af%28y%29+%28x.y%E2%88%88R%29%E4%B8%94f%280%29%E2%89%A00%E4%B8%8D%E7%AD%89%E4%BA%8E%E9%9B%B6%2C%E8%AF%81f%28x%29%E6%98%AF%E5%81%B6%E5%87%BD%E6%95%B0.)
已知函数f(x)满足f(x+y)+f(x-y)=2f(x)(y) (x.y∈R)且f(0)≠0不等于零,证f(x)是偶函数.注:有改动(抱歉)已知函数f(x)满足f(x+y)+f(x-y)=2f(x)*f(y) (x.y∈R)且f(0)≠0不等于零,证f(x)是偶函数.
已知函数f(x)满足f(x+y)+f(x-y)=2f(x)(y) (x.y∈R)且f(0)≠0不等于零,证f(x)是偶函数.
注:有改动(抱歉)
已知函数f(x)满足f(x+y)+f(x-y)=2f(x)*f(y) (x.y∈R)且f(0)≠0不等于零,证f(x)是偶函数.
已知函数f(x)满足f(x+y)+f(x-y)=2f(x)(y) (x.y∈R)且f(0)≠0不等于零,证f(x)是偶函数.注:有改动(抱歉)已知函数f(x)满足f(x+y)+f(x-y)=2f(x)*f(y) (x.y∈R)且f(0)≠0不等于零,证f(x)是偶函数.
证:令X=Y=0,则有F(O)+F(0)=2F(0)的平方
因为f(0)不等于0,所以F(0)=1
再令X=0,则得F(Y)+F(-Y)=2F(0)*F(Y).
因为F(0)=1.
所以F(-Y)=F(Y)
即F(X)为偶函数
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