已知f(x)是定义在(0,+00)上的增函数,且f(x/y)=f(x)-f(y) 求f(1)的值(2) 若f(6)=1,解不等式f(x+3)+f(1除以x)≤2
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![已知f(x)是定义在(0,+00)上的增函数,且f(x/y)=f(x)-f(y) 求f(1)的值(2) 若f(6)=1,解不等式f(x+3)+f(1除以x)≤2](/uploads/image/z/5355786-66-6.jpg?t=%E5%B7%B2%E7%9F%A5f%EF%BC%88x%29%E6%98%AF%E5%AE%9A%E4%B9%89%E5%9C%A8%EF%BC%880%2C%2B00%EF%BC%89%E4%B8%8A%E7%9A%84%E5%A2%9E%E5%87%BD%E6%95%B0%2C%E4%B8%94f%28x%2Fy%29%3Df%28x%EF%BC%89-f%EF%BC%88y%29+%E6%B1%82f%EF%BC%881%EF%BC%89%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89+%E8%8B%A5f%EF%BC%886%EF%BC%89%3D1%2C%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Ff%EF%BC%88x%2B3%EF%BC%89%2Bf%EF%BC%881%E9%99%A4%E4%BB%A5x%EF%BC%89%E2%89%A42)
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