设y=f(x)(x∈R,且x≠0)对任意非零实数x,y都有f(xy)=f(x)+f(y)成立若f(x)在(1,+∞)上单调递增,解不等式f(1/x)-f(2x-1)>=0
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![设y=f(x)(x∈R,且x≠0)对任意非零实数x,y都有f(xy)=f(x)+f(y)成立若f(x)在(1,+∞)上单调递增,解不等式f(1/x)-f(2x-1)>=0](/uploads/image/z/5125269-21-9.jpg?t=%E8%AE%BEy%3Df%28x%29%28x%E2%88%88R%2C%E4%B8%94x%E2%89%A00%29%E5%AF%B9%E4%BB%BB%E6%84%8F%E9%9D%9E%E9%9B%B6%E5%AE%9E%E6%95%B0x%2Cy%E9%83%BD%E6%9C%89f%28xy%29%3Df%28x%29%2Bf%28y%29%E6%88%90%E7%AB%8B%E8%8B%A5f%EF%BC%88x%EF%BC%89%E5%9C%A8%EF%BC%881%2C%2B%E2%88%9E%EF%BC%89%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%2C%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Ff%281%2Fx%29-f%282x-1%29%3E%3D0)
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设y=f(x)(x∈R,且x≠0)对任意非零实数x,y都有f(xy)=f(x)+f(y)成立若f(x)在(1,+∞)上单调递增,解不等式f(1/x)-f(2x-1)>=0
设y=f(x)(x∈R,且x≠0)对任意非零实数x,y都有f(xy)=f(x)+f(y)成立
若f(x)在(1,+∞)上单调递增,解不等式f(1/x)-f(2x-1)>=0
设y=f(x)(x∈R,且x≠0)对任意非零实数x,y都有f(xy)=f(x)+f(y)成立若f(x)在(1,+∞)上单调递增,解不等式f(1/x)-f(2x-1)>=0
f(xy)=f(x)+f(y)
x=y=1,f1=0,
x=y=-1,f(-1)=0
f(-x)=f(-1)+f(x)=f(x) 偶函数
f(1)=fx/1+fx,f1/x=-fx
f(1/x)-f(2x-1)>=0
f(x[2x-1])<=0
-1<=x[2x-1]<=1且x[2x-1]≠0
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