Re: [中學] 今日鳳中教甄
※ 引述《shingai (吸收正能量)》之銘言：
: #2 設a, b, c 為x^3-x^2+2x-3=0 之三根
: 求
: a^3/[(a^2-b^2)(a^2-c^2)]+ b^3/[(b^2-a^2)(b^2-c^2)]+ c^3/[(c^2-b^2)(c^2-a^2)]
: 之值
Let S=a^3/[(a^2-b^2)(a^2-c^2)]+ b^3/[(b^2-a^2)(b^2-c^2)]+
c^3/[(c^2-b^2)(c^2-a^2)]
a^3(b+c) b^3(c+a) c^3(a+b)
Then S*(a+b)(b+c)(c+a)= ----------- + ---------- + --------
(a-b)(a-c) (b-c)(b-a) (c-a)(c-b)
a^3(x-b)(x-c) b^3(x-c)(x-z) c^3(x-a)(x-b)
H(x):= ------------------- + ------------------- + ---------------
(a-b)(a-c) (b-c)(b-a) (c-a)(c-b)
Hence H(x)=x^3 for x=a,b,c.
Since deg(H)<=2, we have H(x)=x^3 - (x-a)(x-b)(x-c).
Note that the coefficient of x in H(x) is the same as -S*(a+b)(b+c)(c+a)
Thus, S=(ab+bc+ca)/[(a+b)(b+c)(c+a)]=-2.
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※ 編輯: motivic (114.45.201.100), 05/03/2015 21:48:59
推
05/03 22:24, , 1F
05/03 22:24, 1F
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