用因式分解法求下列各方程中x的值,答案以 r 和 s 表示
a.x^2 + ( s - r )x - rs=0
b.rx^2 - ( rs + 1)x +s=0
a.用取平方根法求方程(1/r x - s)^2 = 16(其中r =/=0) 中x的值,
b.由此解方程(1/2 x - 開方3)^2=16
一元二次 四題數 要有詳細解答
2007-09-11 1:09 am
回答 (2)
2007-09-11 1:48 am
✔ 最佳答案
a. x^2 + (s - r)x - rs = 0(x + s)(x - r) = 0
x = -s 或 x = r
b. rx^2 - (rs + 1)x + s = 0
(rx - 1)(x - s) = 0
x = 1/r 或 x = s
以上兩題都可以爆開，例如：x^2 + (s - r)x - rs = x^2 + sx - rx - rs = (x + s)x - r(x + s) = (x - r)(x + s)，但直接因式分解比較快。
a. (x/r - s)^2 = 16
x/r - s = 4 或 x/r - s = -4
x = r(s + 4) 或 x = r(s - 4)
b. 設 r = 2，s = √3。
由 a 得 x = 2(√3 + 4) 或 x = 2(√3 - 4)
2007-09-11 1:47 am
1a.x^2 + ( s - r )x - rs=0
(x + s)(x - r) = 0
x = -s or x = r
1b.rx^2 - ( rs + 1)x +s=0
(rx - 1)(x - s) = 0
x = 1/r or x = s
2a. (1/r x - s)^2 = 16
1/r x - s = 4 or 1/r x -s = -4
x = r (4 + s) or x = r (s - 4)
2b. hence, r=2 & s=root 3
thus, x = 2(4 + root3) or x = 2(root3 - 4)
(x + s)(x - r) = 0
x = -s or x = r
1b.rx^2 - ( rs + 1)x +s=0
(rx - 1)(x - s) = 0
x = 1/r or x = s
2a. (1/r x - s)^2 = 16
1/r x - s = 4 or 1/r x -s = -4
x = r (4 + s) or x = r (s - 4)
2b. hence, r=2 & s=root 3
thus, x = 2(4 + root3) or x = 2(root3 - 4)
收錄日期: 2021-04-25 13:28:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070910000051KK02029