HMMT 二月 2007 · TEAM2 赛 · 第 8 题
HMMT February 2007 — TEAM2 Round — Problem 8
题目详情
- [ 20 ] Find the positive real number(s) x such that 3 x − 1 = x − 50 x − 10 x + 25 x + 5 . 2
解析
- [ 20 ] Find the positive real number(s) x such that 3 x − 1 = x − 50 x − 10 x + 25 x + 5 . 2 √ a +2 b − 1 2 2 Answer: 25 + 2 159 . Write a = x − 50 x − 10 and b = x +25 x +5; the given becomes = ab , 2 2 2 so 0 = 2 ab − a − 2 b + 1 = ( a − 1)(2 b − 1). Then a − 1 = x − 50 x − 11 = 0 or 2 b − 1 = 2 x + 50 x + 9 = 0. √ The former has a positive root, x = 25 + 2 159, while the latter cannot, for obvious reasons.