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当x>3时,不等式x+1/(x-1)-a≥0恒成立,求实数a的取值范围
人气:422 ℃ 时间:2020-02-05 03:37:38
解答
解由当x>3时,不等式x+1/(x-1)-a≥0恒成立即当x>3时,不等式x+1/(x-1)≥a恒成立即当x>3时,不等式a≤x+1/(x-1)恒成立即当x>3时,不等式a≤(x-1)+1/(x-1)+1恒成立令f(x)=(x-1)+1/(x-1)+1 (x>3)即a≤f(x)=(x-1...
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