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
• 高! -wxcfan123- ♂ (0 bytes) () 10/11/2023 postreply 18:41:02
• 要證(c^2*b+a^2*c+b^2*a)/abc = 0 還可以這樣做 -wxcfan123- ♂ (91 bytes) () 10/12/2023 postreply 19:16:35
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