F.4 MATHS!

2006-11-27 4:57 am
我想問下..
解下列方程..!
x/2 = 4 - 8/x

x+3/x-1 = 5

開方2x+3 = x

2開方x-3 = x-6

x = 5 - 開方x+1

應該點計啊..-v-''''
thx!X1000000000

回答 (1)

2006-11-27 5:35 am
✔ 最佳答案
1.x/2 = 4 - 8/x
(x/2)2x =(4 - 8/x)2x
x^2=8x-16
x^2-8x+16=0
(x-4)^2=0
x=4

2.x+3/x-1 = 5
(x+3/x-1)x=5x
x^2-x+3=5x
x^2-6x+3=0
x^2-6x+9-6=0
(x-3)^2=6
x=√6+3or3-√6

3.√2x+3=x
2x+3=x^2
x^2-2x-3=0
(x-1)(x+3)=0
x=1or-3

4.2√x-3=x-6
4x-12=x^2-12x+36
x^2-16x+48=0
(x-4)(x-12)=0
x=4or12

5.x=5-√x+1
x-5=-√x+1
x^2-10x+25=x+1
x^2-11x+24=0
x^2-11x+121-97=0
(x-11)^2=97
x=√97+11or11-√97

2006-11-26 21:39:16 補充:
12/x+1 = x - 312+x=x^2-3xx^2-4x-12=0(x-6)(x+2)=0x=6or-2
參考: ...


收錄日期: 2021-04-12 20:51:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061126000051KK05626

檢視 Wayback Machine 備份