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

Question

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

giobas · Answer

=(x-1+x+1)/(x-1)(x+1)=2x/(x^2-1)

saswata4s · Answer

1/(x+1) + 1/(x-1) = (x-1 + x-1)/(x+1)*(x-1) = 2x/(x^2-1); 
Answer D.

Cheers.

BrentGMATPrepNow · Answer

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)}

Answer: D

Mo2men · Answer

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

we left with one answer.

Answer: D

GMATinsight · Answer

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

Answer: Option D

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

GMATinsight · Answer

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

Answer: Option D

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

ScottTargetTestPrep · Answer

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)/

/

2x/(x^2 - 1)

Answer: D

bumpbot · Answer

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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

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