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If x#0 and x#1, and if x is replaced by 1/x everywhere in th

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If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink] New post 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
[Reveal] Spoiler: OA
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink] New post 27 Dec 2012, 05:56
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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

(\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.
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink] New post 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] New post 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] New post 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] New post 30 Jun 2013, 17:47
Does anyone have any similar questions this this one?

I like this problem.

Thanks,
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink] New post 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] New post 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] New post 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] New post 25 Jul 2013, 05:59
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sv3n wrote:
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..


We want to simplify (\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2 so this step seemed quite natural to me.
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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink] New post 25 Jul 2013, 06:10
Tried it several times again. I think I got it know.. thanks.
Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th   [#permalink] 25 Jul 2013, 06:10
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