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If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the [#permalink]
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27 Dec 2012, 05:45
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\((\frac{x+1}{x1})^2\) 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. \((\frac{x+1}{x1})^2\) B. \((\frac{x1}{x+1})^2\) C. \(\frac{x^2+1}{1x^2}\) D. \(\frac{x^21}{x^2+1}\) E. \((\frac{x1}{x+1})^2\)
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Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the [#permalink]
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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 the [#permalink]
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02 May 2013, 14:38
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Bunuel wrote: [b]\((\frac{x+1}{x1})^2\)
\((\frac{\frac{1}{x}+1}{\frac{1}{x}1})^2=(\frac{\frac{1+x}{x}}{\frac{1x}{x}})^2=(\frac{1+x}{1x})^2=(\frac{x+1}{x1})^2\) . How did 1X in the denominator become X1 in last step?
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Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the [#permalink]
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02 May 2013, 20:57
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nikhil007 wrote: Bunuel wrote: [b]\((\frac{x+1}{x1})^2\)
\((\frac{\frac{1}{x}+1}{\frac{1}{x}1})^2=(\frac{\frac{1+x}{x}}{\frac{1x}{x}})^2=(\frac{1+x}{1x})^2=(\frac{x+1}{x1})^2\) . How did 1X in the denominator become X1 in last step? It's not (1x) that became (x1). \((1x)^2\) is simply rewritten as \((x1)^2\). Both \((1x)^2\) and \((x1)^2\) 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 the [#permalink]
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02 May 2013, 21:03
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nikhil007 wrote: Bunuel wrote: [b]\((\frac{x+1}{x1})^2\)
\((\frac{\frac{1}{x}+1}{\frac{1}{x}1})^2=(\frac{\frac{1+x}{x}}{\frac{1x}{x}})^2=(\frac{1+x}{1x})^2=(\frac{x+1}{x1})^2\) . How did 1X in the denominator become X1 in last step? \((1x)^2 = (x1)^2\). For example, \((14)^2 = (41)^2 = 9\). 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 the [#permalink]
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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 the [#permalink]
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25 Jul 2013, 01:48
Bunuel wrote: \((\frac{x+1}{x1})^2\)
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. \((\frac{x+1}{x1})^2\)
B. \((\frac{x1}{x+1})^2\)
C. \(\frac{x^2+1}{1x^2}\)
D. \(\frac{x^21}{x^2+1}\)
E. \((\frac{x1}{x+1})^2\)
\((\frac{\frac{1}{x}+1}{\frac{1}{x}1})^2=(\frac{\frac{1+x}{x}}{\frac{1x}{x}})^2=(\frac{1+x}{1x})^2=(\frac{x+1}{x1})^2\).
Answer: A. I don´t get \((\frac{\frac{1}{x}+1}{\frac{1}{x}1})^2=(\frac{\frac{1+x}{x}}{\frac{1x}{x}})^2\).. 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 the [#permalink]
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25 Jul 2013, 02:12
sv3n wrote: Bunuel wrote: \((\frac{x+1}{x1})^2\)
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. \((\frac{x+1}{x1})^2\)
B. \((\frac{x1}{x+1})^2\)
C. \(\frac{x^2+1}{1x^2}\)
D. \(\frac{x^21}{x^2+1}\)
E. \((\frac{x1}{x+1})^2\)
\((\frac{\frac{1}{x}+1}{\frac{1}{x}1})^2=(\frac{\frac{1+x}{x}}{\frac{1x}{x}})^2=(\frac{1+x}{1x})^2=(\frac{x+1}{x1})^2\).
Answer: A. I don´t get \((\frac{\frac{1}{x}+1}{\frac{1}{x}1})^2=(\frac{\frac{1+x}{x}}{\frac{1x}{x}})^2\).. could you please explain your steps in a few words? Step by step: \((\frac{\frac{1}{x}+1}{\frac{1}{x}1})^2=(\frac{\frac{1+x}{x}}{\frac{1x}{x}})^2\). \((\frac{\frac{1+x}{x}}{\frac{1x}{x}})^2=(\frac{1+x}{x}*\frac{x}{1x})^2\) \((\frac{1+x}{x}*\frac{x}{1x})^2=(\frac{1+x}{1x})^2\) \((\frac{1+x}{1x})^2=(\frac{x+1}{x1})^2\) 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 the [#permalink]
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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 the [#permalink]
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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 the [#permalink]
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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 the [#permalink]
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10 May 2014, 05:50
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Walkabout wrote: \((\frac{x+1}{x1})^2\)
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. \((\frac{x+1}{x1})^2\)
B. \((\frac{x1}{x+1})^2\)
C. \(\frac{x^2+1}{1x^2}\)
D. \(\frac{x^21}{x^2+1}\)
E. \((\frac{x1}{x+1})^2\) 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/1x] ^2 By substituting 2 we get, [(1+2)/(12)] ^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 the [#permalink]
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10 May 2014, 06:19
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nikhil007 wrote: How did 1X in the denominator become X1 in last step? Rewrite the equation so that \(x\) appears as the first term in the equation: \((1x)^2 = (x+1)^2\) let's now rewrite the equation so that \(x\) is positive: \((1x)^2 = (x+1)^2 = [ (1) \cdot (x1)]^2\) the laws of exponents establish that \((a \cdot b)^n = a^n \cdot b^n\) which means that: \((1x)^2 = (x+1)^2 = [ (1) \cdot (x1)]^2 = (1)^2 \cdot (x1)^2\) notice that \((1)^2 = 1 \cdot 1 = 1\) therefore: \((1x)^2 = (x+1)^2 = [ (1) \cdot (x1)]^2 = (1)^2 \cdot (x1)^2 = 1 \cdot (x1)^2 = (x1)^2\)



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Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the [#permalink]
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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 the [#permalink]
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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 the [#permalink]
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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/xy) =(x+y)/(xy) = 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 the [#permalink]
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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 the [#permalink]
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20 Jan 2015, 08:03
This is what I did, but I am not sure if it is correct:
[x+1]^2 / [x1]^2 [(1/x)+1]^2 / [(1/x)1]^2 [(1+x)/x]^2 / [(1x)/x]^2 [(1+x)x]^2 / [(1x)x]^2 (1+x)^2 / (1x)^2 ANS A
It would be easier to read alligned vertically. How do we do that?



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Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the [#permalink]
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13 Feb 2016, 08:26
I replaced x with a value say 2 and then solved the whole problem.Got the answer correct. and its easy too without any confusion.



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Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the [#permalink]
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18 Aug 2016, 16:50
I was tripping over algebra until I realized that no math is needed at all here. You can take your time, triplecheck, and solve this in < 30 seconds.
It's simple algebraic substitution. Here's the key: This formula is valid for all X except 1 and 0. Notice that (A), the correct answer, is exactly the same formula as the question stem.
We could just as easily substitute y, 1/W, pi, z^2, 3^(1/2) or 129,000 in for X and the formula will behave in exactly the same way. All you have to do is doublecheck that you're selecting the equivalent (in this case, identical) formula from the answer choices.
To make this problem more difficult, they could have chosen an answer that is nothing more than an algebraically manipulated version of the original formula. So then, after carefully working through the algebra, you'd wind up with the same thing you started with and still have to massage it a bit in order to compare with the answer choices. I'll be looking out for that trick on GDay.
Someone please correct me if I'm wrong, but that appears to be the shortcut for this problem.




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