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If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
27 Dec 2012, 05:45
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\((\frac{x+1}{x-1})^2\)
If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to
A. \((\frac{x+1}{x-1})^2\)
B. \((\frac{x-1}{x+1})^2\)
C. \(\frac{x^2+1}{1-x^2}\)
D. \(\frac{x^2-1}{x^2+1}\)
E. \(-(\frac{x-1}{x+1})^2\)
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
27 Dec 2012, 05:56
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
02 May 2013, 14:38
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Bunuel wrote:
[b]\((\frac{x+1}{x-1})^2\) \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\) .
How did 1-X in the denominator become X-1 in last step?
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
02 May 2013, 20:57
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nikhil007 wrote:
Bunuel wrote:
[b]\((\frac{x+1}{x-1})^2\) \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\) .
How did 1-X in the denominator become X-1 in last step?
It's not (1-x) that became (x-1).
\((1-x)^2\) is simply rewritten as \((x-1)^2\).
Both \((1-x)^2\) and \((x-1)^2\) are essentially the same
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
02 May 2013, 21:03
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nikhil007 wrote:
Bunuel wrote:
[b]\((\frac{x+1}{x-1})^2\) \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\) .
How did 1-X in the denominator become X-1 in last step?
\((1-x)^2 = (x-1)^2\). For example, \((1-4)^2 = (4-1)^2 = 9\).
The negative sign inside the bracket gets taken care of because of the square.
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
30 Jun 2013, 17:47
Does anyone have any similar questions this this one?
I like this problem.
Thanks,
Hunter
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
25 Jul 2013, 01:48
Bunuel wrote:
\((\frac{x+1}{x-1})^2\) If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to A. \((\frac{x+1}{x-1})^2\) B. \((\frac{x-1}{x+1})^2\) C. \(\frac{x^2+1}{1-x^2}\) D. \(\frac{x^2-1}{x^2+1}\) E. \(-(\frac{x-1}{x+1})^2\) \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\). Answer: A.
I don´t get \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2\).. could you please explain your steps in a few words?
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
25 Jul 2013, 02:12
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sv3n wrote:
Bunuel wrote:
\((\frac{x+1}{x-1})^2\) If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to A. \((\frac{x+1}{x-1})^2\) B. \((\frac{x-1}{x+1})^2\) C. \(\frac{x^2+1}{1-x^2}\) D. \(\frac{x^2-1}{x^2+1}\) E. \(-(\frac{x-1}{x+1})^2\) \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\). Answer: A.
I don´t get \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2\).. could you please explain your steps in a few words?
Step by step:
\((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2\).
\((\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{x}*\frac{x}{1-x})^2\)
\((\frac{1+x}{x}*\frac{x}{1-x})^2=(\frac{1+x}{1-x})^2\)
\((\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\)
Can you please tell me which step didn't you understand?
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
25 Jul 2013, 04:10
I do not understand the first of these four steps. -.- I understand that 1/x+1 is the same as 1+x/x, but why have you done it? I only get it if I see the result and go backwards..
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
25 Jul 2013, 05:59
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
25 Jul 2013, 06:10
Tried it several times again. I think I got it know.. thanks.
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
10 May 2014, 05:50
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Walkabout wrote:
\((\frac{x+1}{x-1})^2\) If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to A. \((\frac{x+1}{x-1})^2\) B. \((\frac{x-1}{x+1})^2\) C. \(\frac{x^2+1}{1-x^2}\) D. \(\frac{x^2-1}{x^2+1}\) E. \(-(\frac{x-1}{x+1})^2\)
We can use a substitution method and process of elimination method to avoid the cumbersome calculations.
Say x = 2
x is replaced by 1/x. So, [(1/x + 1)/(1/x -1)] ^2 = [1+x/1-x] ^2
By substituting 2 we get, [(1+2)/(1-2)] ^2 = 9
Now, substitute x = 2 in the answer choices and an answer choice that gives the final answer as 9 is the correct answer.
Option A gives us 9.
Hence, A is the correct answer.
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
10 May 2014, 06:19
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nikhil007 wrote:
How did 1-X in the denominator become X-1 in last step?
Rewrite the equation so that \(x\) appears as the first term in the equation:
\((1-x)^2 = (-x+1)^2\)
let's now rewrite the equation so that \(x\) is positive:
\((1-x)^2 = (-x+1)^2 = [ (-1) \cdot (x-1)]^2\)
the laws of exponents establish that \((a \cdot b)^n = a^n \cdot b^n\) which means that:
\((1-x)^2 = (-x+1)^2 = [ (-1) \cdot (x-1)]^2 = (-1)^2 \cdot (x-1)^2\)
notice that \((-1)^2 = -1 \cdot -1 = 1\) therefore:
\((1-x)^2 = (-x+1)^2 = [ (-1) \cdot (x-1)]^2 = (-1)^2 \cdot (x-1)^2 = 1 \cdot (x-1)^2 = (x-1)^2\)
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
20 Aug 2014, 13:50
Can't we just multiply the numerator and denominator by x, after substituing in (1/x)?
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
21 Aug 2014, 00:20
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JackSparr0w wrote:
Can't we just multiply the numerator and denominator by x, after substituing in (1/x)?
We require to compute\(\frac{1}{x} + 1\) & \(\frac{1}{x} - 1\) before that
Refer Bunuel's method; done the very best
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If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
09 Dec 2014, 03:15
\((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2\). I just expanded the formula out like this: (1/x+y)(1/x+y)/(1/x+y)(1/x-y) =(x+y)/(x-y) = which is the same as answer choice A when squared (?) Understand the other way mentioned above too but want to check if this is an alternative or if it's incorrect.
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
08 Jan 2015, 19:53
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You can also plug in numbers, such as x=2. Since x is being replaced by (1/x) we will replace x with (1/2). The original equation gives us a solution of 9. plugging in (1/2) into all of the answer solutions present us with A as the only correct answer.
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]
20 Jan 2015, 08:03
This is what I did, but I am not sure if it is correct: [x+1]^2 / [x-1]^2 [(1/x)+1]^2 / [(1/x)-1]^2 [(1+x)/x]^2 / [(1-x)/x]^2 [(1+x)x]^2 / [(1-x)x]^2 (1+x)^2 / (1-x)^2 ANS A It would be easier to read alligned vertically. How do we do that?
Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th
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