已知定义域在区间(0,+∞)上的函数f(x)满足f(x1/x2)=f(x1)-f(x2),且当x>1时,f(x)<0(1)求f(1)的值(2)判断f(x)的单调性(3)若f(3)=-1,解不等式f(|x|)<-2 第三问 为什么是 x>9或x<-9
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![已知定义域在区间(0,+∞)上的函数f(x)满足f(x1/x2)=f(x1)-f(x2),且当x>1时,f(x)<0(1)求f(1)的值(2)判断f(x)的单调性(3)若f(3)=-1,解不等式f(|x|)<-2 第三问 为什么是 x>9或x<-9](/uploads/image/z/3018987-27-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9A%E4%B9%89%E5%9F%9F%E5%9C%A8%E5%8C%BA%E9%97%B4%280%2C%2B%E2%88%9E%29%E4%B8%8A%E7%9A%84%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3f%28x1%2Fx2%29%3Df%28x1%29-f%28x2%29%2C%E4%B8%94%E5%BD%93x%EF%BC%9E1%E6%97%B6%2Cf%28x%29%EF%BC%9C0%EF%BC%881%EF%BC%89%E6%B1%82f%281%29%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89%E5%88%A4%E6%96%ADf%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E6%80%A7%EF%BC%883%EF%BC%89%E8%8B%A5f%283%29%3D-1%2C%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Ff%EF%BC%88%EF%BD%9Cx%EF%BD%9C%29%EF%BC%9C-2+%E7%AC%AC%E4%B8%89%E9%97%AE+%E4%B8%BA%E4%BB%80%E4%B9%88%E6%98%AF+x%EF%BC%9E9%E6%88%96x%EF%BC%9C-9)
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