1x+y1x+y\frac{1}{{x + y}} and 1x−y" role="presentation">1x−y1x−y\frac{1}{{x - y}} is?" /> 1x+y1x+y\frac{1}{{x + y}} and 1x−y" role="presentation">1x−y1x−y\frac{1}{{x - y}} is?" /> 1x+y1x+y\frac{1}{{x + y}} and 1x−y" role="presentation">1x−y1x−y\frac{1}{{x - y}} is?" />

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The sum of $\frac{1}{x + y}$ $\frac{1}{{x + y}}$ and $\frac{1}{x - y}$ $\frac{1}{{x - y}}$ is?

1 $\frac{2 y}{x^{2} - y^{2}}$ $\frac{{2y}}{{{x^2} - {y^2}}}$
2 $\frac{2 x}{x^{2} - y^{2}}$ $\frac{{2x}}{{{x^2} - {y^2}}}$
3 $\frac{- 2 y}{x^{2} - y^{2}}$ $\frac{{ - 2y}}{{{x^2} - {y^2}}}$
4 $\frac{2 x}{y^{2} - x^{2}}$ $\frac{{2x}}{{{y^2} - {x^2}}}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} According to the question, \\ \frac{1}{x + y} + \frac{1}{x - y} \\ = \frac{x - y + x + y}{x^{2} - y^{2}} \\ = \frac{2 x}{x^{2} - y^{2}} \end{aligned}

$\eqalign{ & {\text{According to the question,}} \cr & \frac{1}{{x + y}} + \frac{1}{{x - y}} \cr & = \frac{{x - y + x + y}}{{{x^2} - {y^2}}} \cr & = \frac{{2x}}{{{x^2} - {y^2}}} \cr}$

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and

\frac{1}{x - y}

$\frac{1}{{x - y}}$ is?", "text": "The sum of

\frac{1}{x + y}

$\frac{1}{{x + y}}$ and

\frac{1}{x - y}

$\frac{1}{{x - y}}$ is?", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} According to the question, \\ \frac{1}{x + y} + \frac{1}{x - y} \\ = \frac{x - y + x + y}{x^{2} - y^{2}} \\ = \frac{2 x}{x^{2} - y^{2}} \end{aligned}

$\eqalign{ & {\text{According to the question,}} \cr & \frac{1}{{x + y}} + \frac{1}{{x - y}} \cr & = \frac{{x - y + x + y}}{{{x^2} - {y^2}}} \cr & = \frac{{2x}}{{{x^2} - {y^2}}} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution: