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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, 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

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

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