已知函数f(x)=x^2;+a/x(x≠0,常数a∈R).当a=2时,解不等式f(x)-f(x-1)>2x-1,
人气:466 ℃ 时间:2020-04-04 08:04:23
解答
f(x)=x^2+a/x,当a=2时,有f(x)=x^2+2/x
所以f(x-1)=(x-1)^2+2/x-1
因为f(x)-f(x-1)>2x-1所以x^2+2/x-(x-1)^2+2/x-1>2x-1
化简得-2/x(x-1)>0得-2x(x-1)>0所以0
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