解以下4條方程
6y^2=13y+5
16x^2-40x+25=0
12a^2+5a+3=0
13x+10x^2-3=0
解方程4條
2006-12-05 1:12 am
回答 (3)
2006-12-05 1:23 am
✔ 最佳答案
1. 6y^2=13y+56y^2-13y-5=0
(3y+1)(2y-5)=0
(3y+1)=0 or (2y-5)=0
y= -1/3 or 5/2
2. 16x^2-40x+25=0
(4x-5)^2 =0
x = 5/4
3. 12a^2+5a+3=0
Is it a wrong question? Any typing error?
If it's 12a^2+5a-3=0, the solution will be
(4a+3)(3a-1)=0
a = -3/4 or 1/3
4. 10x^2+13x-3 = 0
(5x-1)(2x+3) = 0
(5x-1)=0 or (2x+3)=0
x = 1/5 or -3/2
2006-12-05 1:51 am
1.6y^2=13y+5
6y^2-13y-5=0
(3y+1)(2y-5)=0
3y+1=0 or 2y-5=0
y= -1/3 or 5/2
2.16x^2-40x+25=0
(4x)^2-40x+5^2=0
(4x-5)^2 =0
4x-5 =0
x = 5/4
4.10x^2+13x-3 = 0
(5x-1)(2x+3) = 0
(5x-1)=0 or (2x+3)=0
x = 1/5 or -3/2
6y^2-13y-5=0
(3y+1)(2y-5)=0
3y+1=0 or 2y-5=0
y= -1/3 or 5/2
2.16x^2-40x+25=0
(4x)^2-40x+5^2=0
(4x-5)^2 =0
4x-5 =0
x = 5/4
4.10x^2+13x-3 = 0
(5x-1)(2x+3) = 0
(5x-1)=0 or (2x+3)=0
x = 1/5 or -3/2
2006-12-05 1:17 am
^ 即係乜？
收錄日期: 2021-04-12 23:04:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061204000051KK02213