解下列各聯立方程
1) (y-5)/2=(x+2y)/3
2x^2+y^2+15=0
2) 5x-2y+1=0
5x^2-y^2=1
數學(聯立方程)
2010-05-20 2:26 am
回答 (2)
2010-05-20 2:44 am
✔ 最佳答案
1) (y-5)/2=(x+2y)/3 2x + 4y = 3y - 15
- (2x + 15) = y , 代入 :
2x^2+y^2+15=0
2x^2 + (- (2x + 15))^2 + 15 = 0
2x^2 + 4x^2 + 60x + 225 + 15 = 0
6x^2 + 60x + 240 = 0
x^2 + 10x + 40 = 0
△ = 10^2 - 4(1)(40) = - 60 < 0
無實數解。
2) 5x-2y+1=0
x = (2y - 1)/5 , 代入 :
5x^2-y^2=1
5[(2y - 1)^2] / 25 - y^2 = 1
(4y^2 - 4y + 1)/5 - y^2 = 1
4y^2 - 4y + 1 - 5y^2 = 5
y^2 + 4y + 4 = 0
(y + 2)^2 = 0
y = - 2
x = (2(-2) - 1)/5
x = - 1
2010-05-23 6:19 am
1) (y-5)/2=(x+2y)/3
2x + 4y = 3y - 15
- (2x + 15) = y , 代入 :
2x^2+y^2+15=0
2x^2 + (- (2x + 15))^2 + 15 = 0
2x^2 + 4x^2 + 60x + 225 + 15 = 0
6x^2 + 60x + 240 = 0
x^2 + 10x + 40 = 0
△ = 10^2 - 4(1)(40) = - 60 < 0
2) 5x-2y+1=0
x = (2y - 1)/5 , 代入 :
5x^2-y^2=1
5[(2y - 1)^2] / 25 - y^2 = 1
(4y^2 - 4y + 1)/5 - y^2 = 1
4y^2 - 4y + 1 - 5y^2 = 5
y^2 + 4y + 4 = 0
(y + 2)^2 = 0
y = - 2
x = (2(-2) - 1)/5
x = - 1
2x + 4y = 3y - 15
- (2x + 15) = y , 代入 :
2x^2+y^2+15=0
2x^2 + (- (2x + 15))^2 + 15 = 0
2x^2 + 4x^2 + 60x + 225 + 15 = 0
6x^2 + 60x + 240 = 0
x^2 + 10x + 40 = 0
△ = 10^2 - 4(1)(40) = - 60 < 0
2) 5x-2y+1=0
x = (2y - 1)/5 , 代入 :
5x^2-y^2=1
5[(2y - 1)^2] / 25 - y^2 = 1
(4y^2 - 4y + 1)/5 - y^2 = 1
4y^2 - 4y + 1 - 5y^2 = 5
y^2 + 4y + 4 = 0
(y + 2)^2 = 0
y = - 2
x = (2(-2) - 1)/5
x = - 1
參考: me
收錄日期: 2021-04-21 22:10:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100519000051KK01024