1. Rationalizing the Denominator

This is a problem that i'm not sure if i finished it or does it still got more steps? thanks

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2. Hello,
Originally Posted by xJenniferx
This is a problem that i'm not sure if i finished it or does it still got more steps? thanks

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$\frac{4}{\sqrt{xy}}\cdot\frac{\sqrt{xy}}{\sqrt{xy} }=\frac{4}{xy}$ ??

I think you forgot $\sqrt{xy}$ in the numerator and you can't go further...

3. so i stop at 4 / xy and forget the (2)(2)/xy?

4. I don't think you needed to reduce 4 to "(2)(2)," but that's right for the most part. You did forget the $\sqrt{xy}$ in the numerator, though.

Have you talked about absolute values yet in class? If so, think of how $\sqrt{x^2}$ isn't exactly the same as $x$. If you haven't gotten that far yet in your class, never mind; you did it just fine.

5. Originally Posted by xJenniferx
so i stop at 4 / xy and forget the (2)(2)/xy?
Yes

But it's $\frac{4 \sqrt{xy}}{xy}$, not $$\frac{4}{xy}$$

6. if we did do absolute value what do i need to change? can't find where i will have to do with x^2

7. Originally Posted by xJenniferx
if we did do absolute value what do i need to change? can't find where i will have to do with x^2
By writing $\sqrt{xy}$ at the very beginning, you already "supposed" xy>0.
But it's always better to write this down

8. EDIT: Moo's right. Haha. Neither $x$ nor $y$ could be negative (technically, both could be negative... but they'd cancel out in the last step), so absolute value signs are redundant.

9. Originally Posted by Nox
It's tricky. Since you multiplied $\sqrt{xy}$ by $\sqrt{xy}$, you'd have $\sqrt{x^2y^2}$, which comes out to | $x$|*| $y$|. My guess is, if you're under 10th or 11th grade or so and/or you haven't gotten to Algebra II, you probably haven't discussed this, and you really needn't worry about it.
no.
It is enough that xy is positive. You don't need to take the absolute values of x and y (though it is not incorrect). I mean it is not useful to write |x| |y|.

10. I know. :P I just didn't think of it before. See the EDIT.