1−11+2–√ + 11−2–√is = ?1−11+2 + 11−2is = ?{\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{is = ?}}" /> 1−11+2–√ + 11−2–√is = ?1−11+2 + 11−2is = ?{\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{is = ?}}" /> 1−11+2–√ + 11−2–√is = ?1−11+2 + 11−2is = ?{\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{is = ?}}" />

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On simplification the value of $1 - \frac{1}{1 + \sqrt{2}} + \frac{1}{1 - \sqrt{2}} is = ?$ ${\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{is = ?}}$

1 $2 \sqrt{2} - 1$ $2\sqrt 2 - 1$
2 $1 - 2 \sqrt{2}$ $1 - 2\sqrt 2$
3 $1 - \sqrt{2}$ $1 - \sqrt 2$
4 $- 2 \sqrt{2}$ $- 2\sqrt 2$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} 1 - \frac{1}{1 + \sqrt{2}} + \frac{1}{1 - \sqrt{2}} \\ = 1 - \frac{(\sqrt{2} - 1)}{(\sqrt{2} + 1) (\sqrt{2} - 1)} - \frac{(\sqrt{2} + 1)}{(\sqrt{2} - 1) (\sqrt{2} + 1)} \\ = 1 - \sqrt{2} + 1 - \sqrt{2} - 1 \\ = 1 - 2 \sqrt{2} \end{aligned}

$\eqalign{ & {\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{ }} \cr & = 1 - \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}} - \frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}} \cr & = 1 - \sqrt 2 + 1 - \sqrt 2 - 1 \cr & = 1 - 2\sqrt 2 \cr}$

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", "text": "On simplification the value of

1 - \frac{1}{1 + \sqrt{2}} + \frac{1}{1 - \sqrt{2}} is  =  ?

${\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{is = ?}}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} 1 - \frac{1}{1 + \sqrt{2}} + \frac{1}{1 - \sqrt{2}} \\ = 1 - \frac{(\sqrt{2} - 1)}{(\sqrt{2} + 1) (\sqrt{2} - 1)} - \frac{(\sqrt{2} + 1)}{(\sqrt{2} - 1) (\sqrt{2} + 1)} \\ = 1 - \sqrt{2} + 1 - \sqrt{2} - 1 \\ = 1 - 2 \sqrt{2} \end{aligned}

$\eqalign{ & {\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{ }} \cr & = 1 - \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}} - \frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}} \cr & = 1 - \sqrt 2 + 1 - \sqrt 2 - 1 \cr & = 1 - 2\sqrt 2 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} 1 - \frac{1}{1 + \sqrt{2}} + \frac{1}{1 - \sqrt{2}} \\ = 1 - \frac{(\sqrt{2} - 1)}{(\sqrt{2} + 1) (\sqrt{2} - 1)} - \frac{(\sqrt{2} + 1)}{(\sqrt{2} - 1) (\sqrt{2} + 1)} \\ = 1 - \sqrt{2} + 1 - \sqrt{2} - 1 \\ = 1 - 2 \sqrt{2} \end{aligned}

$\eqalign{ & {\text{1}} - \frac{1}{{1 + \sqrt 2 }}{\text{ + }}\frac{1}{{1 - \sqrt 2 }}{\text{ }} \cr & = 1 - \frac{{\left( {\sqrt 2 - 1} \right)}}{{\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)}} - \frac{{\left( {\sqrt 2 + 1} \right)}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}} \cr & = 1 - \sqrt 2 + 1 - \sqrt 2 - 1 \cr & = 1 - 2\sqrt 2 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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