已知fx是实数集R上的函数,且对任意x属于R,f(x)=f(x+1)+f(x-1)恒成立,求证fx是周期函数.f(x+2)=f[(x+1)+1]=f(x+1)-f(x) f(x+3)=f[(x+1)+2]=-f[(x+1)-1]=-fx,f(x+6)=f[(x+3)+3]=-f(x+3)=fx
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![已知fx是实数集R上的函数,且对任意x属于R,f(x)=f(x+1)+f(x-1)恒成立,求证fx是周期函数.f(x+2)=f[(x+1)+1]=f(x+1)-f(x) f(x+3)=f[(x+1)+2]=-f[(x+1)-1]=-fx,f(x+6)=f[(x+3)+3]=-f(x+3)=fx](/uploads/image/z/4323325-13-5.jpg?t=%E5%B7%B2%E7%9F%A5fx%E6%98%AF%E5%AE%9E%E6%95%B0%E9%9B%86R%E4%B8%8A%E7%9A%84%E5%87%BD%E6%95%B0%2C%E4%B8%94%E5%AF%B9%E4%BB%BB%E6%84%8Fx%E5%B1%9E%E4%BA%8ER%2Cf%EF%BC%88x%EF%BC%89%3Df%28x%2B1%29%2Bf%28x-1%29%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82%E8%AF%81fx%E6%98%AF%E5%91%A8%E6%9C%9F%E5%87%BD%E6%95%B0.f%28x%2B2%29%3Df%5B%28x%2B1%29%2B1%5D%3Df%28x%2B1%29-f%28x%29+f%28x%2B3%29%3Df%5B%28x%2B1%29%2B2%5D%3D-f%5B%28x%2B1%29-1%5D%3D-fx%2Cf%28x%2B6%29%3Df%5B%28x%2B3%29%2B3%5D%3D-f%28x%2B3%29%3Dfx)
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