解下列各方程
1. ｜3 － 5x｜= 3x － 1
2. x^2 － 2｜x｜ － 3 = 0
F.4 附加數學 - 絕對值 (高手請進)
2007-10-29 3:59 am
回答 (3)
2007-10-29 4:59 am
✔ 最佳答案
1)｜3 － 5x｜= 3x － 1
當x ≧ 3/5
- (3-5x)=3x-1
2x=2
x=1
當x ﹤3/5
3-5x=3x-1
x=1/2（捨去）
2)
x^2 － 2｜x｜ － 3 = 0
｜x｜^2－2｜x｜－3 =0
( ｜x｜+1)(｜x｜－3)=0
｜x｜=3或｜x｜= －1（捨去）
x = ±3
參考: ME^O^””V
2007-10-29 5:16 am
3 － 5x｜= 3x － 1
case(1)x ≧ 3/5
-3+5x=3x-1
2x=2
x=1
in this case have solution x=1
case(2) x﹤3/5
3-5x=3x-1
x=1/2(REJ)
in this case have no solution
SO |3-5x|=3x-1 have solution x=1
2)
case (1) x>=0 case(2)x<0
no solution ~all x are in positive
x^2 - 2|x| - 3 = 0
|x|^2-2|x|-3=0
( |x|+1)(|x|-3)=0
x=+-3 or x= |-1|（rej）
so x^2-2|x|-3=0 have solutions x=3 x=-3
case(1)x ≧ 3/5
-3+5x=3x-1
2x=2
x=1
in this case have solution x=1
case(2) x﹤3/5
3-5x=3x-1
x=1/2(REJ)
in this case have no solution
SO |3-5x|=3x-1 have solution x=1
2)
case (1) x>=0 case(2)x<0
no solution ~all x are in positive
x^2 - 2|x| - 3 = 0
|x|^2-2|x|-3=0
( |x|+1)(|x|-3)=0
x=+-3 or x= |-1|（rej）
so x^2-2|x|-3=0 have solutions x=3 x=-3
2007-10-29 4:04 am
係吾係只係將 絕對值 拆開作 正 負就ok啊....
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