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# Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic

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Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic  [#permalink]

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20 Feb 2017, 01:34
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Given that ∣x∣≠1, the expression $$\frac{1}{(x+1)}+\frac{1}{(x−1)}$$ is equivalent to which of the following?

A. 1/x

B. (2x+1)/(x^2−1)

C. 2x

D. 2x/(x^2−1)

E. 2x^2−1

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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic  [#permalink]

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20 Feb 2017, 01:52
=(x-1+x+1)/(x-1)(x+1)=2x/(x^2-1)
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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic  [#permalink]

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20 Feb 2017, 02:21
1/(x+1) + 1/(x-1) = (x-1 + x-1)/(x+1)*(x-1) = 2x/(x^2-1);

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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic  [#permalink]

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20 Feb 2017, 08:51
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Bunuel wrote:
Given that ∣x∣≠1, the expression $$\frac{1}{(x+1)}+\frac{1}{(x−1)}$$ is equivalent to which of the following?

A. 1/x

B. (2x+1)/(x² - 1)

C. 2x

D. 2x/(x² - 1)

E. 2x² - 1

To combine (add) the two fractions, we need a common denominator.

So, take $$\frac{1}{(x+1)}$$ and create an equivalent fraction by multiplying top and bottom by (x - 1) to get: $$\frac{(x - 1)}{(x² - 1)}$$

ASIDE: (x + 1)(x - 1) = x² - 1

Then take $$\frac{1}{(x-1)}$$ and create an equivalent fraction by multiplying top and bottom by (x + 1) to get: $$\frac{(x + 1)}{(x² - 1)}$$

So, $$\frac{1}{(x+1)}+\frac{1}{(x−1)}$$ = $$\frac{(x - 1)}{(x² - 1)}$$ + $$\frac{(x + 1)}{(x² - 1)}$$

= $$\frac{2x}{(x² - 1)}$$

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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic  [#permalink]

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21 Feb 2017, 02:17
Bunuel wrote:
Given that ∣x∣≠1, the expression $$\frac{1}{(x+1)}+\frac{1}{(x−1)}$$ is equivalent to which of the following?

A. 1/x

B. (2x+1)/(x^2−1)

C. 2x

D. 2x/(x^2−1)

E. 2x^2−1

Another solution Put x =2, then the expression = 1/3 +1 =4/3

We easily can eliminate A, C & E

Apply in B, we get 5/3

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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic  [#permalink]

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21 Feb 2017, 04:11
Bunuel wrote:
Given that ∣x∣≠1, the expression $$\frac{1}{(x+1)}+\frac{1}{(x−1)}$$ is equivalent to which of the following?

A. 1/x

B. (2x+1)/(x^2−1)

C. 2x

D. 2x/(x^2−1)

E. 2x^2−1

Since we have to take the LCM of the two fractions so denominator of the solution much have (x+1)*(x-1) which is (x^2-1) which is available only in Option B and D so Option A, C and E ruled out

Numerator should be the simple addition of (x+1) and (x-1) which is 2x hence

Although substituting the value of x ad checking options also is an easier way but it's always best to solve algeberically till you are stuck and want to substitute values to get out of mess
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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic  [#permalink]

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21 Feb 2017, 04:17
Bunuel wrote:
Given that ∣x∣≠1, the expression $$\frac{1}{(x+1)}+\frac{1}{(x−1)}$$ is equivalent to which of the following?

A. 1/x

B. (2x+1)/(x^2−1)

C. 2x

D. 2x/(x^2−1)

E. 2x^2−1

Since we have to take the LCM of the two fractions so denominator of the solution much have (x+1)*(x-1) which is (x^2-1) which is available only in Option B and D so Option A, C and E ruled out

Numerator should be the simple addition of (x+1) and (x-1) which is 2x hence

Although substituting the value of x ad checking options also is an easier way but it's always best to solve algeberically till you are stuck and want to substitute values to get out of mess
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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic  [#permalink]

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23 Feb 2017, 09:50
Bunuel wrote:
Given that ∣x∣≠1, the expression $$\frac{1}{(x+1)}+\frac{1}{(x−1)}$$ is equivalent to which of the following?

A. 1/x

B. (2x+1)/(x^2−1)

C. 2x

D. 2x/(x^2−1)

E. 2x^2−1

Let’s first get a common denominator of (x + 1)(x -1) by multiplying the first expression by (x -1)/(x -1) and the second expression by (x + 1)/(x + 1):

(x -1)/[(x -1)(x +1)] + (x + 1)/[(x -1)(x+1)]

[(x - 1) + (x + 1)]/[(x -1)(x + 1)]

2x/(x^2 - 1)

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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic &nbs [#permalink] 23 Feb 2017, 09:50
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