※ 引述《Rfaseven11 (覺醒哥)》之銘言：
: QQ
: https://i.imgur.com/ZOQCJfs.jpg
1. f(x) = (x-a)(x-b)(x-c) = (x-a)[(x-b)(x-c)] = (x-a)[x*x - (b+c)*x + bc]
f'(x) = (x-b)(x-c) + (x-a)[2x - (b+c)]
f'(a) = (a-b)(a-c), f'(b) = (b-a)(b-c), f'(c) = (c-a)(c-b)

chx64的1/2悖論真的很經典呢