2011年“北约”自主招生数学压轴题(第七题):求f(x)=|x-1|+|2x-1|+…+|2011x-1|的最小值.文【1】给出了一种解法,本文立足绝对值定义,给出一种较简单的解法.分析由于各项是绝对值,故解题思路可按绝对值的定义去掉绝对值符号切入.当1/(k+1)≤x≤1/k时,有kx-1≤0,(k+1)x-1≥0则函数f(x)=|x-1|+|2x-1|+…+ 2011 “NATO ” self-enrollment math finale (seventh): Find f (x) = | x-1 | + | 2x-1 | + ... + | 2011x-1 | ] Gives a solution, this article based on the definition of absolute value, gives a simple solution.As the analysis of the absolute value, so the problem-solving ideas can be defined as the absolute value of the symbol to remove the absolute cut.When 1 / (k + 1) ≤x≤1 / k, kx-1≤0, (k + 1) x-1≥0 then the function f (x) = | x-1 | + | 2x-1 | + +