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July 9th, 2016, 08:51 PM   #1
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simplifying expressions with negative exponents

This is the third time I'm going to ask here and I'm very glad I'm embracing fun times math can give. Still, I have a lot of things to learn like the one above in the title . SO we all know that if we have a negative exponent example a^-2 it will be equivalent to 1/a^2. So what if we are going have equation like this. x(1-2x)^-3/2+(1-2x)^-1/2 how are we going to simplify this thing .... according to the book of Precal in pre-requisite section we will have x/(1-2x)^3/2+1/(1-2x)^1/2...SO according to the book I have red .. to combine the expression you need to get the LCD and by doing so you need to remove the common factor with the smaller exponent..which will give us 1-x/(1-2x)^3/2 ......................so I was like "What the heck did just happen".............. Can you please elaborate...also it is also stated that when factoring we need to subtract exponent.....why subtraction ..why not addition ...
so in summary this is my question..
A.How did we arrive at that answer? please elucidate in a detailed way please.
B.Why do we need to subtract the exponents, why not add them ?
Thank you in advance
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July 9th, 2016, 10:14 PM   #2
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I'll address the question of simplifying $\dfrac{x}{(1 - 2x)^{3/2}} + \dfrac{1}{(1 - 2x)^{1/2}}$ first.

The first step is to get a common denominator. Note that $\dfrac{1}{(1 - 2x)^{1/2}} = \dfrac{1 - 2x}{(1 - 2x)^{3/2}}$, so our expression becomes
$$\dfrac{x}{(1 - 2x)^{3/2}} + \dfrac{1 - 2x}{(1 - 2x)^{3/2}} = \dfrac{1 - x}{(1 - 2x)^{3/2}}$$



As for your question about factorisation,say I wanted to factor the expression $a^2 + a^3$.
Obviously, the HCF (highest common factor) is $a^2$, so the factorisation would be
$$a^2(\dfrac{a^2}{a^2} + \dfrac{a^3}{a^2}) = a^2(a^{2 - 2} + a^{3 - 2}) = a^2(1 + a)$$
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July 10th, 2016, 01:10 AM   #3
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Azzajazz has combined several steps in his working.

I recommend beginners writing out every simple single step until they achieve sufficient familiarity to combine steps.


So to explain

Take

$\displaystyle \frac{1}{(1-2x)^{\frac{1}{2}}}$

Multiply by 1 (that's right multiply by one)

$\displaystyle \frac{1}{(1-2x)^{\frac{1}{2}}}*1$

But

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

So we have

$\displaystyle \frac{1}{(1-2x)^{\frac{1}{2}}}*\frac{1-2x}{1-2x}$

Which equals

$\displaystyle \frac{(1-2x)}{(1-2x)^{\frac{3}{2}}}$

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July 10th, 2016, 10:07 PM   #4
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Of course, the difficulty is knowing you need to multiply by $\dfrac{1 - 2x}{1 - 2x}$. This is simply a matter of experience.
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