先找出一個合理的“最小搭配”

注意到7個男生3個女生的平均分恰好是63

同比變換

就可以知道

男生是70人雞兔同籠評課,女生30人

男生比女生多40人
古代的雞兔同" />

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歡迎您訪問雞兔同籠評課,雞兔同籠問題!

雞兔同籠評課,雞兔同籠問題

更新時間:2021-06-13 00:00:34作者:admin2

這個么


先找出一個合理的“最小搭配”


注意到7個男生3個女生的平均分恰好是63


同比變換


就可以知道


男生是70人雞兔同籠評課,女生30人


男生比女生多40人

古代的雞兔同籠問題

《孫子算經》
今有雉兔同籠,上有三十五頭,下有九十四足,問雉兔各幾何?(有若干只雞兔同在一個籠子里,從上面數,有35個頭;從下面數,有94只腳。求籠中各有幾只雞和兔?)
解:
2×35=70(只)
94-70=24 (只)
24÷2=12 (只)
35-12=23(只)

LZ要的使這個不咯?
解題思路:
先假設它們全是雞,于是根據雞兔的總數就可以算出在假設下共有幾只腳,把這樣得到的腳數與題中給出的腳數相比較,看看差多少,每差2只腳就說明有1只兔,將所差的腳數除以2,就可以算出共有多少只兔。
概括起來,解雞兔同籠題的基本關系式是:兔數=(實際腳數-每只雞腳數×雞兔總數)÷(每只兔子腳數-每只雞腳數)。類似的,也可以假設全是兔子。
列方程的方法:設兔子的數量為X,雞的數量為Y
那么:X+Y=35那么4X+2Y=94 這個算方程解出后得:兔子有12只,雞有23只。

還要詳細的話
去百度百科吧
O(∩_∩)O~

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