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f(x)>2k-2恒成立即f(x)-2k+2>0恒成立則x²;-2x+1-k²;-2k+2>0恒成立x∈(0,+無窮)首先找到二次函數的對稱軸-2a分之b=1又函數影像開口向上可知當x=1時取到最小值若x∈(0,+無窮)x²;-2x+1-k²;-2k…
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