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If x > 0, then 1/ = 1/sqrt3x 1/ 1/(xsqrt2)

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Manager
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Joined: 10 Oct 2005
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If x > 0, then 1/ = 1/sqrt3x 1/ 1/(xsqrt2) [#permalink]

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New post 18 Oct 2005, 20:24
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

1/sqrt3x

1/[2sqrt(2x)]

1/(xsqrt2)

(sqrt2-1)/sqrtx

(1+2)/sqrtx

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Director
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Re: p/s sqrt [#permalink]

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New post 18 Oct 2005, 20:28
trickygmat wrote:
If x > 0, then 1/sqrt(2x)+sqrtx=

1/sqrt3x

1/[2sqrt(2x)]

1/(xsqrt2)

(sqrt2-1)/sqrtx

(1+2)/sqrtx


your parentheses are ambiguous :!:

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Director
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 [#permalink]

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New post 18 Oct 2005, 20:28
D (sqrt2-1)/sqrtx

(1/sq(2x)+sq(x))*([sq(2x)-sq(x)]/[sq(2x)+sq(x)])
= [sq(2x)-sq(x)]/x = sq(x)*[sq(2)-1]/x = [sq(2)-1]/sq(x)

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Director
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 [#permalink]

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New post 18 Oct 2005, 21:18
gsr wrote:
D (sqrt2-1)/sqrtx

(1/sq(2x)+sq(x))*([sq(2x)-sq(x)]/[sq(2x)+sq(x)])
= [sq(2x)-sq(x)]/x = sq(x)*[sq(2)-1]/x = [sq(2)-1]/sq(x)


can you mutiply by

(sq(2x) - sq(x))
(sq(2x) +sq(x)) ?

how do you get from

sq(x)[sq(2) - 1] -------> [sq(2)-1]
--------- X ------------------- Sq(X)


Just asking...

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Director
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 [#permalink]

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New post 18 Oct 2005, 21:23
Titleist wrote:
gsr wrote:
D (sqrt2-1)/sqrtx

(1/sq(2x)+sq(x))*([sq(2x)-sq(x)]/[sq(2x)+sq(x)])
= [sq(2x)-sq(x)]/x = sq(x)*[sq(2)-1]/x = [sq(2)-1]/sq(x)


can you mutiply by

(sq(2x) - sq(x))
(sq(2x) +sq(x)) ?

how do you get from

sq(x)[sq(2) - 1] -------> [sq(2)-1]
--------- X ------------------- Sq(X)

Just asking...


sorry typo...multiply Numerator and Den. by sqrt(2x)-sqrtx

on the other one -

sq(x)*[sq(2)-1]/x = sq(x)*[sq(2)-1]/[sq(x)*sq(x)]

= [sq(2)-1]/sq(x)

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Director
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 [#permalink]

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New post 18 Oct 2005, 21:28
gsr wrote:
Titleist wrote:
gsr wrote:
D (sqrt2-1)/sqrtx

(1/sq(2x)+sq(x))*([sq(2x)-sq(x)]/[sq(2x)+sq(x)])
= [sq(2x)-sq(x)]/x = sq(x)*[sq(2)-1]/x = [sq(2)-1]/sq(x)


can you mutiply by

(sq(2x) - sq(x))
(sq(2x) +sq(x)) ?

how do you get from

sq(x)[sq(2) - 1] -------> [sq(2)-1]
--------- X ------------------- Sq(X)

Just asking...


sorry typo...multiply Numerator and Den. by sqrt(2x)-sqrtx

on the other one -

sq(x)*[sq(2)-1]/x = sq(x)*[sq(2)-1]/[sq(x)*sq(x)]

= [sq(2)-1]/sq(x)


ahhh yes. this can mean only two things; i'm getting tired and my new avatar is distracting the hell out of me. i'm going to bed goodnight all! see you tomorrow. :sleep:

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Manager
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 [#permalink]

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New post 19 Oct 2005, 04:47
Finally I got it!!! This question was driving me crazy.

There is another typo:

Stem is 1/(sq(2x)+sq(x)) not (1/sq(2x)+sq(x)),

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 [#permalink]

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New post 19 Oct 2005, 05:51
I have edited the question for trickygmat so the parentheses are correctly placed.

By the way, D is my answer as well. Same method as the rest.

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VP
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 [#permalink]

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New post 19 Oct 2005, 16:47
1/[sqrt(2x)+sqrt(x)]= 1/ ( (sqrt(2) + 1)sqrt(x) ) .

Since (sqrt(2) -1 )(sqrt(2) + 1 ) = 2 - 1 = 1. Put that in the numerator and we get (sqrt(2) -1 )/sqrt(x).

All this sqrt text intead of the symbol makes typing miserable :).

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  [#permalink] 19 Oct 2005, 16:47
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If x > 0, then 1/ = 1/sqrt3x 1/ 1/(xsqrt2)

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