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Kdel425
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Danou
\(\sqrt{2} +\frac{1}{2 + \sqrt{2}} + \frac{1}{\sqrt{2} - 2} = \)

A. \(0\)

B. \(2\)

C. \(\frac{\sqrt{2}}{2}\)

D. \(\sqrt{2}\)

E. \(2\sqrt{2}\)




 
\(\frac{1}{2 + \sqrt{2}} + \frac{1}{\sqrt{2} - 2}\)

\(\frac{\sqrt{2} - 2 + 2 + \sqrt{2}}{(2 + \sqrt{2})(\sqrt{2} - 2)}\)­

\(\frac{2\sqrt{2}}{2-4}\)­

\(\frac{-2\sqrt{2}}{2}\)­

\(-\sqrt{2}\)­

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

\(\sqrt{2} -\sqrt{2}\)

= 0

Option A

 ­
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Goal here is to create common denominators.

Remember conjugates with roots in denominators. Multiply by opposite. Ex: 1/x + sqrt y should be multiplied by sqrt y -x.

leave the sqrt 2 for now move on the second term and then the third term and multiply by conjugates.

You should get sqrt 2 + (2- srt2) / 2 - (sqrt2 +2)/2

Multiply the second and third term by 2 and get 2 sqrt 2 + 2 - sqrt 2 - sqrt 2 + 2

Everything washes answer is 0

A
Kdel425
\(\sqrt{2} + \frac{1}{2+\sqrt{2}} + \frac{1}{\sqrt{2}-2} = \)

A. 0

B. 2

C. \(\frac{\sqrt{2}}{2}\)

D. \(\sqrt{2}\)

E. ­\(2\sqrt{2}\)­

Attachment:
Screenshot (5).png
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