Ok, I have a math test in 3 hours and I can't for the life of me figure out how to simplify complex fractions such as these.

Simplify the expression:

3(1+x)^1/3 - x(1+x)^-2/3 <- numerator

(1+x)^2/3 <- denominator

Anyone out there able to help a desperate guy?
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Last edited by Burtonjp at May 17, 2011,
wolframalpha.com
Make sure the entire numerator is in parentheses, the fractional powers have their own set of parentheses

(1-x^2)^(-1/2)

and see what comes up.
Last edited by Jerry.thewise at May 17, 2011,
Yeah sure... So (3-3x)/2 + (x^2 - X^4)^-1/2
thats what you should be getting if you multiply your brackets first. I'm guessing that's right unless I've read the equation wrong. Get back to me
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Quote by DaMarsbarPerson
Yeah sure... So (3-3x)/2 + (x^2 - X^4)^-1/2
thats what you should be getting if you multiply your brackets first. I'm guessing that's right unless I've read the equation wrong. Get back to me

Its simplifying complex fractional expressions

This is the fraction

3(1-x)^1/2 + x^2(1-x^2)^-1/2 <-----------numerator
__________________________
1-x^2 <-----------denominator
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to the power of 1/2 is square root so i think
/3-3x+(-/2x^2-/X^4)
/3-/3x-2x-x^2
thats as far as i got
/=square root
Good luck
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Quote by Stugg334

Right hang on. I've read it all wrong...
If you multiply it out you'll get
3 x Sq.root (1-x) + x^2 over (1-x)

I think
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What year are you in TS?

I fear/loathe the day I end up learning this stuff
Are you sure it isn't this? Numerator and denominator are each in their own set of brackets

[3*((1-x^2)^(1/2)) + (x^2)*((1-x^2)^(-1/2)) ] / [ (1-x^2) ]

The thing that changed was the first term, how it has an x^2 rather than just an x. This would make the problem much easier to simplify.
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Feign illness
Ah, I'm an idiot.... I molded two different questions together while copying it down.. haha

OK

this is the REAL question

3(1+x)^1/3 - x(1+x)^-2/3 <- numerator

(1+x)^2/3 <- denominator

What year are you in TS? I fear/loathe the day I end up learning this stuff

Basic question from start of precalc... It SHOULDN'T be difficult... I think my brains just fried today.
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Last edited by Burtonjp at May 17, 2011,
Really, the problem you gave us cannot be "simplified" any more than it is. You can split it up at the plus sign into two fractions
[3*((1-x)^(1/2)) ]/ [ (1-x^2) ] + [ (x^2)*((1-x^2)^(-1/2)) ] / [ (1-x^2) ]
but then you can't do anything to the first term that is bold.

If it were as I had said,
[3*((1-x^2)^(1/2)) + (x^2)*((1-x^2)^(-1/2)) ] / [ (1-x^2) ]

[3*((1-x^2)^(1/2)) ]/ [ (1-x^2) ] + [(x^2)*((1-x^2)^(-1/2)) ] / [ (1-x^2) ]

3 * [ (1-x^2)^(-1/2) ] + [ (x^2) ] * [ (1-x^2)^(-3/2) ]

Remember that the negative exponents mean that those terms could be rewritten as the denominator of a fraction

EDIT
well forget that
Quote by Jerry.thewise
Are you sure it isn't this? Numerator and denominator are each in their own set of brackets

[3*((1-x^2)^(1/2)) + (x^2)*((1-x^2)^(-1/2)) ] / [ (1-x^2) ]

The thing that changed was the first term, how it has an x^2 rather than just an x. This would make the problem much easier to simplify.
+1 to this.

If not...

Expand (1-x^2)^(1/2) to (1-x)^(-1/2)(1+x)^(-1/2)

Then you can take the whole (1-x)^(-1/2) out of the brackets on the numerator, giving you:

(1-x)^(-1/2)[3(1-x)+x^2(1+x)^(-1/2)]

Then you can take that (1-x)^(-1/2) to the denominator to make it (1-x^2)(1-x)^(1/2).

After that maybe take out the (1+x)^(-1/2) and move that to the denominator too. Don't think there's too much you can do with it after that.
3(1-x)^1/2 + x^2(1-x^2)^-1/2
__________________________
1-x^2

(1-x^2) = (1-x)(1+x)
so (1-x^2)^-1/2 = {(1-x)(1+x)}^-1/2

3(1-x)^1/2 + x^2 * (1-x)^-1/2 * (1+x)^-1/2
____________________________________
1-x^2

3(1-x)^-1/2 =
(1-x)^-1/2 = (1-x) * (1-x)^-3/2
(1+x)^-1/2 = (1+x) * (1+x)^-3/2

putting that all back in.

3 * (1-x)^-3/2 + x^2 * (1-x)^-3/2 * (1+x)^-3/2
________________ ______________________________________
(1+x)

then cancel

3 + X^2
_____________ _____________
(1+x)(1-x)^3/2 (1-x^2)^3/2

=

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

that probably went wrong somewhere and you might be able to simplify it further, but thats what I got from ****ing about with it while bored.

Edit: The forum mushed it all together, where there is a space in the fractions or a + on the numerator that means 2 separate fractions
Last edited by Nat400 at May 17, 2011,

2x+3 <-(numerator)
(x+1)^4/3 <-(denominator)
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Quote by Burtonjp
Ah, I'm an idiot.... I molded two different questions together while copying it down.. haha

OK

this is the REAL question

3(1+x)^1/3 - x(1+x)^-2/3 <- numerator

(1+x)^2/3 <- denominator

Take (1+x)^(-2/3) out of the brackets on the numerator to give

(1+x)^(-2/3)[3(x+1)-x]

Then take it to the denominator as (1+x)^2/3 . Giving you (1+x)^4/3.

Leaving 3(x+1)-x = 2x +3 on the numerator.

Edit:
Quote by Burtonjp

2x+3 <-(numerator)
(x+1)^4/3 <-(denominator)
Yep, pretty sure this is right.
Last edited by 12345abcd3 at May 17, 2011,
[3(1+x)^1/3 - x(1+x)^-2/3] / [(1+x)^2/3]

split into two terms

[3(1+x)^1/3] / [(1+x)^2/3] - [x(1+x)^-2/3] / [(1+x)^2/3]

Here's a cool trick that can help you see what is really going on

let u=1+x

[3(u)^1/3] / [(u)^2/3] - [x(u)^-2/3] / [(u)^2/3]

Fix up the terms u so that you only have one u for each term. Combine them by taking the u in the numerator and its exponent and subtracting the exponent of the u in the denominator. For the first term, (1/3)-(2/3)=(-1/3). Because this is negative, the u will end up in the denominator, or you can leave the exponent negative in the numerator.

3(u)^(-1/3) - x(u)^(-4/3)

Looks pretty simple here, so lets put that 1+x back in wherever we have u

3(1+x)^(-1/3) - x(1+x)^(-4/3)

Quote by 12345abcd3
Take (1+x)^(-2/3) out of the brackets on the numerator to give

(1+x)^(-2/3)[3(x+1)-x]

Then take it to the denominator as (1+x)^2/3 . Giving you (1+x)^4/3.

Leaving 3(x+1)-x = 2x +3 on the numerator.

Edit:
Yep, pretty sure this is right.

I like this answer better now though
Last edited by Jerry.thewise at May 17, 2011,
Alrighty, thanks for the help.... brain clicked on, I remembered how to solve these, and I figured it out. Thanks for all the quick responses.
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