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HMMT 十一月 2011 · GEN 赛 · 第 5 题

HMMT November 2011 — GEN Round — Problem 5

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [ 5 ] Find all real values of x for which 1 1 1 √ + √ = . √ √ 4 x + x − 2 x + 2 + x
解析
  1. [ 5 ] Find all real values of x for which 1 1 1 √ √ √ + √ = . 4 x + x − 2 x + 2 + x 257 Answer: We note that 16 1 1 1 √ √ = + √ √ 4 x + x − 2 x + 2 + x √ √ √ √ x − x − 2 x + 2 − x = √ √ + √ √ √ √ √ √ ( x + x − 2)( x − x − 2) ( x + 2 + x )( x + 2 − x ) √ √ √ √ x − x − 2 x + 2 − x = + 2 2 √ √ 1 = ( x + 2 − x − 2) , 2 so that √ √ 2 x + 2 − 2 x − 2 = 1 . Squaring, we get that √ √ 8 x − 8 ( x + 2)( x − 2) = 1 ⇒ 8 x − 1 = 8 ( x + 2)( x − 2) . Squaring again gives 2 2 64 x − 16 x + 1 = 64 x − 256 , 257 so we get that x = . 16