find the real root(s)

2008-03-06 6:57 am
6/x-2 =x+3

(4/x+1)(3-x )=3

x^6+7x^3 -8=0

x-2= square root 2x-1

logx+log(x-3)=1

log (2x-10)-2logx =-1

9^x(x+1)=27^x+1

回答 (2)

2008-03-06 7:39 am
✔ 最佳答案
As follows~

圖片參考:http://i248.photobucket.com/albums/gg191/ncy_0916/log-1.jpg?t=1204731516


2008-03-05 23:40:32 補充:
log M + log N = log(MN)
log M - log N = log(M/N)
log10 = 1

2008-03-06 23:47:09 補充:
第4題的x ﹣2 = √(2x ﹣1)
x求到1 或 5
但個1要rejected,因為
L.H.S. = 1﹣2
= -1
R.H.S. = √(2 ﹣1)
= 1
即是L.H.S.≠R.H.S>
2008-03-06 7:31 am
6/x-2 =x+3

(4/x+1)(3-x )=3

multiply x for both sides on above 2 questions


x^6+7x^3 -8=0

Try let x^3 = y, it becomes a quadratic equation


x-2= square root 2x-1

Take square on both sides


logx+log(x-3)=1

log(x^2-3x) = 1

x^2-3x=10
....


log (2x-10)-2logx =-1
try the method like the above question~


add oil!!
hope can help you to do the maths, except the last question..
參考: me


收錄日期: 2021-04-14 20:21:58
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080305000051KK03127

檢視 Wayback Machine 備份