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3^(1/2)/3 + 1/(3 + 3^(1/2)) + 1/(3^(1/2) - 3)

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3^(1/2)/3 + 1/(3 + 3^(1/2)) + 1/(3^(1/2) - 3) [#permalink]

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New post 30 Mar 2017, 06:29
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

68% (01:45) correct 32% (01:32) wrong based on 127 sessions

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Re: 3^(1/2)/3 + 1/(3 + 3^(1/2)) + 1/(3^(1/2) - 3) [#permalink]

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New post 30 Mar 2017, 06:41
\(\frac{\sqrt{3}}{3} + \frac{1}{3+\sqrt{3}} + \frac{1}{\sqrt{3}-3}\)
= \(\frac{\sqrt{3}}{3} + \frac{1}{3+\sqrt{3}} - \frac{1}{3-\sqrt{3}}\)
= \(\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}\)
= 0

Answer: C
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Re: 3^(1/2)/3 + 1/(3 + 3^(1/2)) + 1/(3^(1/2) - 3) [#permalink]

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New post 30 Mar 2017, 06:46
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Bunuel wrote:
\(\frac{\sqrt{3}}{3} + \frac{1}{3+\sqrt{3}} + \frac{1}{\sqrt{3}-3} =\)


A. \(\sqrt{3}\)

B. 1

C. 0

D. -1

E. \(-\sqrt{3}\)


Hi

Rationalize and Simplify \(\frac{1}{3+\sqrt{3}} and \frac{1}{\sqrt{3}-3}\)

We get \((2sqrt{3}+{3-\sqrt{3}}-{\sqrt{3}+3})/6\)

By Simplification we get 0
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Re: 3^(1/2)/3 + 1/(3 + 3^(1/2)) + 1/(3^(1/2) - 3) [#permalink]

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New post 30 Mar 2017, 07:07
Vyshak wrote:
\(\frac{\sqrt{3}}{3} + \frac{1}{3+\sqrt{3}} + \frac{1}{\sqrt{3}-3}\)
= \(\frac{\sqrt{3}}{3} + \frac{1}{3+\sqrt{3}} - \frac{1}{3-\sqrt{3}}\)
=] [m]\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}[/m
= 0

Answer: C


hi
vayshak ,although i got the answer but can please explain how did you come with this [m]\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}[/m
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3^(1/2)/3 + 1/(3 + 3^(1/2)) + 1/(3^(1/2) - 3) [#permalink]

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New post 30 Mar 2017, 07:18
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nks2611 wrote:
Vyshak wrote:
\(\frac{\sqrt{3}}{3} + \frac{1}{3+\sqrt{3}} + \frac{1}{\sqrt{3}-3}\)
= \(\frac{\sqrt{3}}{3} + \frac{1}{3+\sqrt{3}} - \frac{1}{3-\sqrt{3}}\)
= \(\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}\)
= 0

Answer: C


hi
vayshak ,although i got the answer but can please explain how did you come with this \(\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}\)


\(\frac{1}{3+\sqrt{3}} - \frac{1}{3-\sqrt{3}} = \frac{3-\sqrt{3}}{(3+\sqrt{3})(3-\sqrt{3})} - \frac{3+\sqrt{3}}{(3+\sqrt{3})(3-\sqrt{3})} = \frac{(3-\sqrt{3}) - (3+\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})} = \frac{-2\sqrt{3}}{3^2-\sqrt{3}^2} = -\frac{2\sqrt{3}}{6}=-\frac{\sqrt{3}}{3}\)
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3^(1/2)/3 + 1/(3 + 3^(1/2)) + 1/(3^(1/2) - 3) [#permalink]

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New post 12 May 2017, 17:28
nks2611 wrote:
Vyshak wrote:
\(\frac{\sqrt{3}}{3} + \frac{1}{3+\sqrt{3}} + \frac{1}{\sqrt{3}-3}\)
= \(\frac{\sqrt{3}}{3} + \frac{1}{3+\sqrt{3}} - \frac{1}{3-\sqrt{3}}\)
=] [m]\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}[/m
= 0

Answer: C


hi
vayshak ,although i got the answer but can please explain how did you come with this [m]\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}[/m


It's important not to forget that there's an assumed 1 in front of the radical - sometimes members post solutions without putting a 1 in front of the radical - which can be confusing for those less familiar with math or perhaps sometimes those forgetting- that or sometimes when you cross multiply a fraction and figure that the numerators will be the same people don't write out the answer because there's minds work a bit faster- which is good for the GMAT maybe just a bit more work for those learning math or rusty.
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Re: 3^(1/2)/3 + 1/(3 + 3^(1/2)) + 1/(3^(1/2) - 3) [#permalink]

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New post 15 May 2017, 08:27
A. 3√3

B. 1

C. 0 :twisted:

D. -1

E. −3√

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Re: 3^(1/2)/3 + 1/(3 + 3^(1/2)) + 1/(3^(1/2) - 3)   [#permalink] 15 May 2017, 08:27
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