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

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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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

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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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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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.

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:

\((1-x)^2 = (-x+1)^2 = [ (-1) \cdot (x-1)]^2 = (-1)^2 \cdot (x-1)^2 = 1 \cdot (x-1)^2 = (x-1)^2\)

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.

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?

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.

I was tripping over algebra until I realized that no math is needed at all here. You can take your time, triple-check, 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 double-check 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 G-Day.

Someone please correct me if I'm wrong, but that appears to be the shortcut for this problem.

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: \((\frac{1}{x+1/1/x-1})^2=(\frac{1+x}{x/1-x/x})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\)

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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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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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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I get the feeling that you are actually correct here. And it is quite a clever point. We could replace x with anything and of course the identical expression (option A) will give us the same value.