Re: [中學] 等差數列
※ 引述《oxs77 (安)》之銘言：
: A,B,C,D,E 為等差數列
: A^3+6C^3+E^3 (A+E) [(A+E)^2-3AE] +6C^3 (2C) [(2C)^2-3AE] + 6C^3
: 求 ------------ = -------------------------- = ----------------------------
: B^3+D^3 (B+D) [(B+D)^2-3BD] (2C) [(2C)^2-3BD]
4C^2 -3AE +3C^2 7C^2 - 3(C-2d)(C+2d)
= ------------------ = ------------------------
4C^2 -3BD 4C^2 - 3(C-d)(C+d)
4C^2 -12d^2
= ---------------= 4
C^2 -3d^2
秒殺法 取 A=B=C=D=E=1 ==> 4
--
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 1.169.43.200
推
03/05 21:31, , 1F
03/05 21:31, 1F
推
03/05 21:36, , 2F
03/05 21:36, 2F
推
03/05 21:42, , 3F
03/05 21:42, 3F
→
03/05 21:46, , 4F
03/05 21:46, 4F
討論串 (同標題文章)
