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Can you make these 2 fractions into 1?

$$2\sqrt{9-2x} - \dfrac{2x}{\sqrt{9-2x}}$$

I thought you could make them into $ \dfrac{-2x+18}{\sqrt{9-2x}}$

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    rather $18-4x-2x$ in the nominator2012-10-13
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    You're right, thank you.2012-10-13
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    Else it's maximally ok2012-10-13

1 Answers 1

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You're almost correct.

\begin{align} 2\sqrt{9-2x} - \frac{2x}{\sqrt{9-2x}} &= 2\sqrt{9-2x} \times \frac{\sqrt{9-2x}}{\sqrt{9-2x}} - \frac{2x}{\sqrt{9-2x}}\\ &= \frac{2(\sqrt{9-2x})^2}{\sqrt{9-2x}} - \frac{2x}{\sqrt{9-2x}}\\ &= \frac{2(9-2x)}{\sqrt{9-2x}} - \frac{2x}{\sqrt{9-2x}}\\ &= \frac{2(9-2x) - 2x}{\sqrt{9-2x}}\\ &= \frac{18 - 4x - 2x}{\sqrt{9-2x}}\\ &= \frac{18-6x}{\sqrt{9-2x}} \end{align}

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    I assume that we should make the statement that $x$ is not equal 9/2.2012-10-13
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    That is implicit as soon as the initial expression is given. In fact, you need $x < \frac{9}{2}$.2012-10-13
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    OK, thanks, may be the OP should take notice.2012-10-13