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

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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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11 Jan 2017, 19:40
Is this problem testing the fact that multiplying or dividing both the numerator and the denominator of a fraction by the same number does not change the value of the fraction? In that case, a quick glance would show that the answer is A.
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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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15 Feb 2017, 03:39
1
For me the key to solving this question was to understand that $$(1-x)^2=(x-1)^2$$. To check that this is correct I tested a few cases: $$(5-3)^2=4, (3-5)^2=4; (1-6)^2=25, (6-1)^2=25$$.

Full solution: $$(\frac{1}{x+1/1/x-1})^2=(\frac{1+x}{x/1-x/x})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2$$
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Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the  [#permalink]

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09 Apr 2017, 21:38
Bunuel wrote:
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. $$-(\frac{x-1}{x+1})^2$$

$$(\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2$$.

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?

Hi Bunuel,

Regarding the last step: $${ \left( \cfrac { 1+x }{ 1-x } \right) }^{ 2 }={ \left( \cfrac { x+1 }{ x-1 } \right) }^{ 2 }$$

The reason why we can move the x and 1 in the numerator is because they're positive, so it's indifferent which one comes first, correct?
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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 Apr 2017, 22:53
diegocml wrote:
Bunuel wrote:
sv3n wrote:
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?

Hi Bunuel,

Regarding the last step: $${ \left( \cfrac { 1+x }{ 1-x } \right) }^{ 2 }={ \left( \cfrac { x+1 }{ x-1 } \right) }^{ 2 }$$

The reason why we can move the x and 1 in the numerator is because they're positive, so it's indifferent which one comes first, correct?

No.

For the numerator: changing the order of the addends does not change the sum: a + b = b + a. So, 1 + x = x + 1.

For the denominator: (a - b)^2 = (b - a)^2, so (1 - x)^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 the  [#permalink]

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24 Aug 2017, 06:33
Hi,
I have a query if someone can explain.
I get ans as (1+x/1-x)^2. Options which confuse me are
B) and E)

Ans 1+x/1-x if both numerator and denominator is multiplied by - then we get (x-1/x+1) (B) and negative cancel out
or -(x-1/x+1)(E) negative stays... I am not sure. Can we multiply by a -ve sign when there is a bracket with a 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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06 Sep 2017, 06:14
$$(\frac{x+1}{x-1})^2$$

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. $$-(\frac{x-1}{x+1})^2$$

Hey there - I have been struggling understanding questions in this format. Is there a guide anywhere for the same? Or is there a better browser to be used to read questions with fractions and powers?
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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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06 Sep 2017, 06:19
zoompastthisGMAT 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. $$-(\frac{x-1}{x+1})^2$$

Hey there - I have been struggling understanding questions in this format. Is there a guide anywhere for the same? Or is there a better browser to be used to read questions with fractions and powers?

Check Algebra chapter in Ultimate GMAT Quantitative Megathread.

Hope it helps.
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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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12 Oct 2018, 22:24
VeritasKarishma could you please share your approach here specially how it becomes from (1+x/1-x)^2 to (x+1/x-1)^2 ? 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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16 May 2019, 07:01
$$(\frac{x+1}{x-1})^2$$

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. $$-(\frac{x-1}{x+1})^2$$

Another quick hack to solve this type of question is via number substitution which makes it 0

Put X=-1 in all the options and you would only option A as same

Since X#0 & X#1 , we can X= -1

now when it is replaced by 1/X essentially it is still -1, Only Option A is giving ans as 0 ( Sames as question equation)

Correct me if I am wrong
Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the   [#permalink] 16 May 2019, 07:01

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