若定义在R上的函数f(x)对任意的x1,x2∈R,都有f(x1+x2)=f(x1)+f(x2)-1成立,且当x〉0时,f(x)〉1.(1)求证:f(x1)-1为奇函数(2)求证:f(x)是R上的增函数(3)若f(4)=5,解不等式f(3m的2次方-m-2)〈3
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![若定义在R上的函数f(x)对任意的x1,x2∈R,都有f(x1+x2)=f(x1)+f(x2)-1成立,且当x〉0时,f(x)〉1.(1)求证:f(x1)-1为奇函数(2)求证:f(x)是R上的增函数(3)若f(4)=5,解不等式f(3m的2次方-m-2)〈3](/uploads/image/z/2667290-50-0.jpg?t=%E8%8B%A5%E5%AE%9A%E4%B9%89%E5%9C%A8R%E4%B8%8A%E7%9A%84%E5%87%BD%E6%95%B0f%28x%29%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84x1%2Cx2%E2%88%88R%2C%E9%83%BD%E6%9C%89f%28x1%2Bx2%29%3Df%28x1%29%2Bf%28x2%29-1%E6%88%90%E7%AB%8B%2C%E4%B8%94%E5%BD%93x%E3%80%890%E6%97%B6%2Cf%28x%29%E3%80%891.%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9Af%28x1%29-1%E4%B8%BA%E5%A5%87%E5%87%BD%E6%95%B0%EF%BC%882%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9Af%28x%29%E6%98%AFR%E4%B8%8A%E7%9A%84%E5%A2%9E%E5%87%BD%E6%95%B0%EF%BC%883%EF%BC%89%E8%8B%A5f%284%29%3D5%2C%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8Ff%283m%E7%9A%842%E6%AC%A1%E6%96%B9-m-2%29%E3%80%883)
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