還有一簡單解法. 因三數不同 故圴不為0
1/a +2/b+3/c->a+b+c+c/a+a/b+b/c=3
然後證明C/a+a/b+b/c=0
C/a+a/b+b/c=(c^2*b+a^2*c+b^2*a)/abc
1*b+2*c+3*a and 1+2+3 -> c^2*b+a^2*c+b^2*a=0
還有一簡單解法. 因三數不同 故圴不為0
1/a +2/b+3/c->a+b+c+c/a+a/b+b/c=3
然後證明C/a+a/b+b/c=0
C/a+a/b+b/c=(c^2*b+a^2*c+b^2*a)/abc
1*b+2*c+3*a and 1+2+3 -> c^2*b+a^2*c+b^2*a=0
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高!
-wxcfan123-
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10/11/2023 postreply
18:41:02
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要證(c^2*b+a^2*c+b^2*a)/abc = 0 還可以這樣做
-wxcfan123-
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10/12/2023 postreply
19:16:35
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