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lrh2006 Posted 2017-7-1 09:56 |Read mode
这道题目看到一些不同的答案,请各位指点迷津,谢谢! P6PF][6FTXFMSU`O@)6QU.png

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 Author| lrh2006 Posted 2017-7-1 23:52
请各位帮帮我吧,不胜感激

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realnumber Posted 2017-7-2 08:28
$c_n=$max{$a_n,b_n$},观察y=-x+t与$y=3^{x-3}$的图像,
因为$c_3$最小,那么有$c_4\ge c_3$即$b_4\ge a_3$($b_4\ge b_3$显然成立,$b_4\ge a_3 \ge a_4$)
同样$a_2\ge b_3$($a_2\ge a_3$也显然成立).
这样解得$3\le t \le 6$.

好像没有没考虑到什么了.

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 Author| lrh2006 Posted 2017-7-2 09:28
回复 3# realnumber

嗯嗯,我也觉得这样很清楚了,谢谢噢!

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