己知:(b^2+c^2-a^2)/2bc+(c^2+a^2-b^2)/2ca+(a^2+b^2-c^2)/2ab=1,
己知:(b^2+c^2-a^2)/2bc+(c^2+a^2-b^2)/2ca+(a^2+b^2-c^2)/2ab=1,
求[(b^2+c^2-a^2)/(2bc)]^2009+[(c^2+a^2-b^2)/(2ca)]^2009+[(a^2+b^2-c^2)/(2ab)]^2009的值
求[(b^2+c^2-a^2)/(2bc)]^2009+[(c^2+a^2-b^2)/(2ca)]^2009+[(a^2+b^2-c^2)/(2ab)]^2009的值
数学人气:499 ℃时间:2020-04-09 17:19:33
优质解答
解;依题意的.(b^2+c^2-a^2)/2bc+(c^2+a^2-b^2)/2ca+(a^2+b^2-c^2)/2ab=1 则[(b^2+c^2-a^2)/(2bc)]^2009+[(c^2+a^2-b^2)/(2ca)]^2009+[(a^2+b^2-c^2)/(2ab)]^2009 =1这是一条推断题,只有分析好题意你就会发现其中的...过程~~谢谢= =
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