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最佳答案:f'(x) = [a(x + 1) - (ax - 1)]/(x + 1)² = (a + 1)/(x + 1)² < 0 (x < -1时)a < -1a的确
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最佳答案:f′(x)=-a/[2(a-1)√(3-ax)]∵f(x)在区间【0,1】上是减函数∴f′(x)0∴-a/(a-1)0∴a