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

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Math Expert
Joined: 02 Sep 2009
Posts: 58434
3^(1/2)/3 + 1/(3 + 3^(1/2)) + 1/(3^(1/2) - 3)  [#permalink]

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30 Mar 2017, 07:29
10
00:00

Difficulty:

35% (medium)

Question Stats:

74% (02:06) correct 26% (02:24) wrong based on 193 sessions

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$$\frac{\sqrt{3}}{3} + \frac{1}{3+\sqrt{3}} + \frac{1}{\sqrt{3}-3} =$$

A. $$\sqrt{3}$$

B. 1

C. 0

D. -1

E. $$-\sqrt{3}$$

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Posts: 1684
Location: India
Re: 3^(1/2)/3 + 1/(3 + 3^(1/2)) + 1/(3^(1/2) - 3)  [#permalink]

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30 Mar 2017, 07:41
1
$$\frac{\sqrt{3}}{3} + \frac{1}{3+\sqrt{3}} + \frac{1}{\sqrt{3}-3}$$
= $$\frac{\sqrt{3}}{3} + \frac{1}{3+\sqrt{3}} - \frac{1}{3-\sqrt{3}}$$
= $$\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}$$
= 0

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

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30 Mar 2017, 07:46
1
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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30 Mar 2017, 08:07
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

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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30 Mar 2017, 08:18
1
1
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

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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12 May 2017, 18:28
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

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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15 May 2017, 09:27
A. 3√3

B. 1

C. 0

D. -1

E. −3√

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

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09 Oct 2018, 17:49
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Re: 3^(1/2)/3 + 1/(3 + 3^(1/2)) + 1/(3^(1/2) - 3)   [#permalink] 09 Oct 2018, 17:49
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