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# If x > 0, then 1/ = A. 1/v(3x) B. 1/ C. 1/(xv2) D.

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Manager
Joined: 14 Jan 2006
Posts: 96

Kudos [?]: 40 [0], given: 2

Schools: HKUST
If x > 0, then 1/ = A. 1/v(3x) B. 1/ C. 1/(xv2) D. [#permalink]

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23 Sep 2008, 13:55
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x > 0, then 1/[v(2x)+vx] =

A. 1/v(3x)
B. 1/[2v(2x)]
C. 1/(xv2)
D. (v2-1)/vx
E. (1+v2)/vx

Kudos [?]: 40 [0], given: 2

GMAT Tutor
Joined: 24 Jun 2008
Posts: 1339

Kudos [?]: 1954 [0], given: 6

Re: GMAT SET 19 - 2 [#permalink]

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23 Sep 2008, 16:06
I assume you mean the question to read:

$$\frac{1}{\sqrt{x} + \sqrt{2x}} = \text{ ?}$$

If so, we have:

$$\frac{1}{\sqrt{x} + \sqrt{2x}$$

$$= \frac{1}{\sqrt{x} + \sqrt{2} \sqrt{x}$$

$$= \frac{1}{\sqrt{x}(1 + \sqrt{2})}$$

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

$$= \frac{1 - \sqrt{2}}{- \sqrt{x} }$$

$$= \frac{ \sqrt{2} - 1}{\sqrt{x}}$$

Normally roots aren't written in denominators, so that would normally be written as:

$$\frac{ \sqrt{2x} - \sqrt{x} }{x}$$
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Kudos [?]: 1954 [0], given: 6

VP
Joined: 05 Jul 2008
Posts: 1402

Kudos [?]: 437 [0], given: 1

Re: GMAT SET 19 - 2 [#permalink]

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23 Sep 2008, 17:43
nikhilpoddar wrote:
If x > 0, then 1/[v(2x)+vx] =

A. 1/v(3x)
B. 1/[2v(2x)]
C. 1/(xv2)
D. (v2-1)/vx
E. (1+v2)/vx

NP,

How about leaving an idea to people that v represents sqrt . If we take it as letter v, it can be some thing else. I was wondering why the Q is so simple, until I saw Ian's interpretation

Kudos [?]: 437 [0], given: 1

Re: GMAT SET 19 - 2   [#permalink] 23 Sep 2008, 17:43
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