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I have been working on this problem for a while and can't quite figure out how to do it. How can I turn this expression into a single quotient?

$\frac{2x}{(1-x)^\frac12} + \frac{2(1-x)^\frac{-1}{2}}{3}$

The first thing I did was move the negative exponent down in the denominator to obtain this expression.

$\frac{2x}{(1-x)^\frac12} + \frac{2}{3(1-x)^\frac{1}{2}}$

However, I'm stuck on how to combine it as a single quotient because the fractional exponents are throwing me off.

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    $\begingroup$ Just factor out the $1/(1-x)^{1/2}$ from both terms. $\endgroup$ – Robert Israel Aug 24 '17 at 0:51
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Hint: it may be more obvious if you work with $(1 - x)^{-1/2}$ in the numerators of the fractions.

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You can treat the fractional exponent in exactly the same way you would if the exponent were one or even zero, to see why consider factoring it out.

Consider this as $2x(1-x)^{\frac{-1}{2}}+\frac{2}{3}(1-x)^{\frac{-1}{2}}$, then simply factor out $(1-x)^{\frac{-1}{2}}$ to obtain $(1-x)^{\frac{-1}{2}}(2x+\frac{2}{3})$. This is clearly $(1-x)^{\frac{-1}{2}}(\frac{6x+2}{3})$, then multiply it out to obtain one fraction, we have:

$\frac{6x+2}{3\sqrt{1-x}}=\frac{6x+2}{3(1-x)^{\frac{1}{2}}}$

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