Mathematical Problem

2007-08-13 5:14 am
1.在nxn的方格紙中,把部分小方格塗成紅色,然後劃去其中2行和2列,無論怎樣劃,都至少有一個紅色的小方格沒有被劃去,則至少要塗多少小方格?
(In terms of n)

2.36x^4+48x^3+88x^2+48x+36=0
x=?

回答 (2)

2007-08-13 9:42 pm
✔ 最佳答案
1.
n ≦ 2 , 題目不成立

n = 3 - -> 塗9 格(全塗)

n = 4 - -> 塗6 格(塗對角線四格, 另二角各一格)

n ≧ 5 - ->塗5 格(塗對角線上, 任選五格)
2.
題誤?88x^2 - -> 84x^2
36x^4+48x^3+84x^2+48x+36=0

3x^4 + 4x^3 + 7x^2 + 4x + 3 = 0
3x^4 + 3x^3 + 3x^2 + x^3 + x^2 + x + 3x^2 + 3x + 3 = 0

(x^2+ x + 1)(3x^2 + x + 3) = 0

(x^2+ x + 1) = 0, x = - (1/2) + (√3)i / 2 or - (1/2) - (√3)i / 2

(3x^2 + x + 3) = 0, x = - (1/6) + (√35)i / 6 or - (1/6) - (√35)i / 6
2007-08-21 9:58 am
1. 唔識

2. Solve 36x^4 + 48x³ + 88x² + 48x + 36 = 0

Luckily, this equation is a symmetric equation. We can solve this type of equation by 'x + 1/x' method.

36x^4 + 48x³ + 88x² + 48x + 36 = 0
9x² + 12x + 22 + 12/x + 9/x² = 0
9(x² + 1/x²) + 12(x + 1/x) + 22 = 0
9(x² + 2 + 1/x²) – 18 + 12(x + 1/x) + 22 = 0
9(x + 1/x)² + 12(x + 1/x) + 4 = 0
(3(x + 1/x) + 2)² = 0
3x + 2 + 3/x = 0
3x² + 2x + 3 = 0
x = [– 2 ± √(2² – 4(3)(3))]/(2 × 3)
x = (– 2 ± √– 32)/6
x = (– 2 ± (4√2)i)/6
x = (– 1 + (2√2)i)/3 (repeated root) or (– 1 – (2√2)i)/3 (repeated root)


收錄日期: 2021-04-18 22:54:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070812000051KK05047

檢視 Wayback Machine 備份