函数f(x)满足关系f(xy)=f(x)+f(y)(x,y属于R) (1)求f(1) (2)求f(⅓)+f(&f函数f(x)满足关系f(xy)=f(x)+f(y)(x,y属于R)(1)求f(1)(2)求f(⅓)+f(½)+f(1)+f(2)+
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![函数f(x)满足关系f(xy)=f(x)+f(y)(x,y属于R) (1)求f(1) (2)求f(⅓)+f(&f函数f(x)满足关系f(xy)=f(x)+f(y)(x,y属于R)(1)求f(1)(2)求f(⅓)+f(½)+f(1)+f(2)+](/uploads/image/z/5293391-23-1.jpg?t=%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E6%BB%A1%E8%B6%B3%E5%85%B3%E7%B3%BBf%EF%BC%88xy%EF%BC%89%3Df%EF%BC%88x%29%2Bf%28y%29%EF%BC%88x%2Cy%E5%B1%9E%E4%BA%8ER%EF%BC%89+%281%EF%BC%89%E6%B1%82f%EF%BC%881%EF%BC%89+%EF%BC%882%EF%BC%89%E6%B1%82f%EF%BC%88%26%238531%3B%EF%BC%89%2Bf%EF%BC%88%26f%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E6%BB%A1%E8%B6%B3%E5%85%B3%E7%B3%BBf%EF%BC%88xy%EF%BC%89%3Df%EF%BC%88x%29%2Bf%28y%29%EF%BC%88x%2Cy%E5%B1%9E%E4%BA%8ER%EF%BC%89%281%EF%BC%89%E6%B1%82f%EF%BC%881%EF%BC%89%EF%BC%882%EF%BC%89%E6%B1%82f%EF%BC%88%26%238531%3B%EF%BC%89%2Bf%EF%BC%88%26frac12%3B%EF%BC%89%2Bf%EF%BC%881%EF%BC%89%2Bf%EF%BC%882%EF%BC%89%2B)
函数f(x)满足关系f(xy)=f(x)+f(y)(x,y属于R) (1)求f(1) (2)求f(⅓)+f(&f函数f(x)满足关系f(xy)=f(x)+f(y)(x,y属于R)(1)求f(1)(2)求f(⅓)+f(½)+f(1)+f(2)+
函数f(x)满足关系f(xy)=f(x)+f(y)(x,y属于R) (1)求f(1) (2)求f(⅓)+f(&f
函数f(x)满足关系f(xy)=f(x)+f(y)(x,y属于R)
(1)求f(1)
(2)求f(⅓)+f(½)+f(1)+f(2)+f(3)
函数f(x)满足关系f(xy)=f(x)+f(y)(x,y属于R) (1)求f(1) (2)求f(⅓)+f(&f函数f(x)满足关系f(xy)=f(x)+f(y)(x,y属于R)(1)求f(1)(2)求f(⅓)+f(½)+f(1)+f(2)+
因为函数f(x)满足关系f(xy)=f(x)+f(y)
所以f(0×1)=f(0)+f(1)即f(0)=f(0)+f(1)
所以f(1)=0,
f(1/3)+f(3)=f(1)=0
f(1/2)+f(2)=f(1)=0
所以f(⅓)+f(½)+f(1)+f(2)+f(3) =0
(1)令x=1,y=1带入易得f(1)=0
(2)令x=a,y=a分之1带入可得f(a)+f(a分之一)=0,所以原式=0
第一问 令x、y=1就可以得出 f(1)=0
第二问f(1/3)+f(3)+f(1)+f(2)+f(1/2)
=f(1)+f(1)+f(1)
=0
(1)为0 (2)为0
(1)令x=1,y=1,则f(1)=f(1)+f(1),即f(1)=0
(2)f(1/3)+f(1/2)+f(1)+f(2)+f(3)=[f(1/3)+f(3)]+f(1)+[f(2)+f(1/2)]=f(1)f(1)f(1)=0.
f(1/3)+f(3)=f[(1/3)*3]=f(1)