[解題] 國一下 數學 比與比例
1.年級：國一
2.科目：數學
3.章節：考卷、第二冊比與比例
4.題目：已知a、b、c皆不為0， 5a 5b 5c
____ = ____ = _____ = k
b+c c+a a+b
若a+b+c=0，求k??
5.想法：
想請問以下解法哪步開始錯了：
5a=k(b+c)
5b=k(c+a)
5c=k(a+b) (因為a+b+c=0，所以從這三式其中一式知道Ex：5a=-ak，所以k=-5
但想知道這種解法哪裡開始錯...)
三式相加得5(a+b+c)=k(2a+2b+2c)
→ 5(a+b+c)=2k(a+b+c)
因為a+b+c=0不能直接約分，所以移項得5(a+b+c)-2k(a+b+c)=0
因式分解得 (5-2k)(a+b+c)=0
因a+b+c=0 故得k為任意數
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※ 編輯: VaLenTi1007 來自: 59.112.226.33 (09/13 01:06)
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