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设tanx=3tany(0
人气:170 ℃ 时间:2020-02-21 23:23:04
解答
tan(x-y)=(tanx-tany)/(1+tanxtany)
=2tany/(1+3tan^2y)
=2/(1/tany+3tany)
≤2/2√(1/tany*3tany)
=1/√3
=√3/3
=tanπ/6
x-y的最大值为:π/6
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