數學知識交流---解方程(4)

(1) 解：x + ( x + 1 ) + ( x + 2 ) + ... + ( x + x ) = 117180
x (x+1) + (1 + 2 + ... + x) = 117180
x^2 + x + x(x+1)/2 = 117180
2x^2 + 2x + x^2 + x = 234360
3x^2 + 3x - 234360 = 0
x^2 + x - 78120 = 0
(x + 280) (x - 279) = 0
所以x1 = -280, x2 = 279

(2) 解：x - ( x - 1 ) - ( x - 2 ) - ... - ( x - x ) = -11935
x - x^2 + (1 + 2 + ... + x) + 11935 = 0
x - x^2 + x(x+1)/2 + 11935 = 0
2x - 2x^2 + x^2 + x + 23870 = 0
x^2 - 3x - 23870 = 0
x1 ≈ 156, x2 ≈ -153

2011-06-07 13:23:35 補充：
(2) 解：x - ( x - 1 ) - ( x - 2 ) - ... - ( x - x ) = -11627
x - x^2 + (1 + 2 + ... + x) + 11627 = 0
x - x^2 + x(x+1)/2 + 11627 = 0
2x - 2x^2 + x^2 + x + 23254 = 0
x^2 - 3x - 23254 = 0
(x - 154) (x + 151) = 0
x1 = 154, x2 = -151