己知函数f（x）=|2|x|﹣1|．（Ⅰ）求不等式f（x）≤1的解集A；（Ⅱ）当m，n∈A时，证明：|m+n|≤mn+1．

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己知函数f（x）=|2|x|﹣1|．

（Ⅰ）求不等式f（x）≤1的解集A；

（Ⅱ）当m，n∈A时，证明：|m+n|≤mn+1．

上一题下一题 0.0难度选择题更新时间:2019-07-13 01:57:36

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（Ⅰ）解：f（x）≤1即|2|x|﹣1|≤1．

∴﹣1≤2|x|﹣1≤1，∴|x|≤1

解得：﹣1≤x≤1，所以A=﹣1，1

（Ⅱ）证明：要证：|m+n|≤mn+1，即证（m+n）²≤（mn+1）²

因为（m+n）²﹣（mn+1）²=m²+n²﹣m²n²﹣1=（m²

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