高中数学解题过程求解释请写出过程:已知a>0,函数f(x)=(x/(a+1)-1)^2+(a/x-1)^2-2/3(x>0)1、当a=1/3时,求函数f(x)的最小值. (2)证明:当a >= 1/5时,函数f(x)有零点解:(1)令3x/4=m,1/(3x)=n 则mn=1/4 y=(m-1)^2
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![高中数学解题过程求解释请写出过程:已知a>0,函数f(x)=(x/(a+1)-1)^2+(a/x-1)^2-2/3(x>0)1、当a=1/3时,求函数f(x)的最小值. (2)证明:当a >= 1/5时,函数f(x)有零点解:(1)令3x/4=m,1/(3x)=n 则mn=1/4 y=(m-1)^2](/uploads/image/z/8183328-24-8.jpg?t=%E9%AB%98%E4%B8%AD%E6%95%B0%E5%AD%A6%E8%A7%A3%E9%A2%98%E8%BF%87%E7%A8%8B%E6%B1%82%E8%A7%A3%E9%87%8A%E8%AF%B7%E5%86%99%E5%87%BA%E8%BF%87%E7%A8%8B%3A%E5%B7%B2%E7%9F%A5a%3E0%2C%E5%87%BD%E6%95%B0f%28x%29%3D%28x%2F%28a%2B1%29-1%29%5E2%2B%28a%2Fx-1%29%5E2-2%2F3%28x%3E0%291%E3%80%81%E5%BD%93a%3D1%2F3%E6%97%B6%2C%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC.+%EF%BC%882%EF%BC%89%E8%AF%81%E6%98%8E%EF%BC%9A%E5%BD%93a+%3E%3D+1%2F5%E6%97%B6%2C%E5%87%BD%E6%95%B0f%28x%29%E6%9C%89%E9%9B%B6%E7%82%B9%E8%A7%A3%EF%BC%9A%281%29%E4%BB%A43x%2F4%3Dm%2C1%2F%283x%29%3Dn+++++%E5%88%99mn%3D1%2F4+++++y%3D%28m-1%29%5E2)
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