1/(1-x)<1/(x-5)
2006-08-09 03:24:01 · 13 answers · asked by Daphne Loo 1 in Science & Mathematics ➔ Mathematics
i get the answer was diferent with my friend, so i want to make sure who's answer is correct!
2006-08-09 03:36:08 · update #1
1/(1-x) < 1/(x-5)
1/(1-x) - 1/(x-5) < 0
expend the above the equation:
(2x - 6) / [ (1-x) (x-5) ] < 0 ---- equation (1)
segregate equation (1) into 2 parts:
Thus: (2x - 6) < 0 --- equation (2)
(1-x) (x-5) --- equation (3)
equation (3) cannot be equal to zero in order for equation (1) to be valid, thus equation (3) need to be greater than zero or less than zero.
the answer: 1< x < 3 and x > 5
2006-08-10 17:16:09 · answer #1 · answered by AJ 2 · 1⤊ 0⤋
x<3
2006-08-09 06:48:02 · answer #2 · answered by Lewiy 3 · 0⤊ 0⤋
The two previous answers are correct: x < 3
Think about it this way...
1/(1-x) is smaller than 1/(x-5), so the (1-x) part must be greater than the (x-5) part...
1-x > x-5 ... (1+5) > x+x ... 6 > 2x, x < 3
2006-08-09 03:47:51 · answer #3 · answered by David R 3 · 1⤊ 0⤋
x>0
2006-08-09 03:29:20 · answer #4 · answered by Michael 5 · 0⤊ 0⤋
See the table below
x1/(1-x) 1/(x-5) 1/(1-x)<1/(x-5)
1inf -0.25 No
2-1 -0.333333333Yes
3-0.5 -0.5 No
4-0.333333333-1 Yes
5-0.25 inf Yes
6-0.2 1 Yes
7-0.1666666670.5 Yes
8-0.1428571430.333333333Yes
9-0.125 0.25 Yes
10-0.1111111110.2 Yes
11-0.1 0.166666667Yes
12-0.0909090910.142857143Yes
13-0.0833333330.125 Yes
14-0.0769230770.111111111Yes
15-0.0714285710.1 Yes
16-0.0666666670.090909091Yes
17-0.0625 0.083333333Yes
18-0.0588235290.076923077Yes
2006-08-09 18:25:42 · answer #5 · answered by Clinkit 2 · 0⤊ 0⤋
1 / (1 - x) < 1 / (x - 5)
So:
x - 5 < 1 - x
Rearranging gives you x < 3, not valid at x = 1
2006-08-09 03:27:46 · answer #6 · answered by Status: Paranoia 4 · 0⤊ 0⤋
cases: (1-x) < 0; (1-x) = 0; (1-x) > 0; (x-5) < 0; (x-5) = 0; (x-5) > 0
Intervals:
(-infinity, 1), (1, 3), (3, 5), (5, +infinity)
The inequality is true for the intervals (1, 3) and [5, +infinity).
2006-08-09 03:29:26 · answer #7 · answered by AlphaOne_ 5 · 0⤊ 0⤋
because its absolute value it can have an excellent or detrimental value and nonetheless proceed to be functional. so for you to make 2 inequalities. |3x + 5| > 10 ----> 3x + 5 > 10 and the different is -(3x + 5) > 10 3x + 5 > 10 3x>5 x>5/3 -(3x + 5) > 10 -3x-5 > 10 -3x>15 x<-5 so now you position both solutions at the same time. x>5/3 or x<-5
2016-11-23 17:35:56 · answer #8 · answered by ? 4 · 0⤊ 0⤋
1/(1-x)<1/(x-5)=
2006-08-09 04:08:38
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answer #9
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answered by Victor C 3
·
0⤊
0⤋
1-x>x-5=
6-x>x
6>2x
3>x,
1<(1-x)/(x-5)=
1<-5+6x-x²
6<6x+x²
0
x-5<1-x
2x<6
x<3
2006-08-09 03:35:27 · answer #10 · answered by ag_iitkgp 7 · 0⤊ 0⤋