若函数f(x)=loga(2x^2+x) (a>0,a≠1)在区间(1/2,1)内恒有f(x)>0,则f(x)的单调递增区间是————?我自己做的答案是负无穷到负二分一~可是错了.
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![若函数f(x)=loga(2x^2+x) (a>0,a≠1)在区间(1/2,1)内恒有f(x)>0,则f(x)的单调递增区间是————?我自己做的答案是负无穷到负二分一~可是错了.](/uploads/image/z/1170430-70-0.jpg?t=%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%29%3Dloga%282x%5E2%2Bx%29+%28a%3E0%2Ca%E2%89%A01%29%E5%9C%A8%E5%8C%BA%E9%97%B4%281%2F2%2C1%29%E5%86%85%E6%81%92%E6%9C%89f%28x%29%3E0%2C%E5%88%99f%28x%29%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%E6%98%AF%E2%80%94%E2%80%94%E2%80%94%E2%80%94%3F%E6%88%91%E8%87%AA%E5%B7%B1%E5%81%9A%E7%9A%84%E7%AD%94%E6%A1%88%E6%98%AF%E8%B4%9F%E6%97%A0%E7%A9%B7%E5%88%B0%E8%B4%9F%E4%BA%8C%E5%88%86%E4%B8%80%7E%E5%8F%AF%E6%98%AF%E9%94%99%E4%BA%86.)
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