Check out some similar questions!

adkloup · Jun 9, 2009, 07:48 AM

for all x>0 and y>0, the radical expression
√x
3√x - √y

is equivalent to :

3x - √y
9x + y

3x - √xy
3x + y

3x + √xy
9x - y

3x + √xy
3x - y

x
3x - y

those are spost to be fractions didn't know how to do that

Perito · Jun 9, 2009, 08:44 AM

Originally Posted by adkloup

for all x>0 and y>0, the radical expression
√x
3√x - √y

is equivalent to :

3x - √y
9x + y

3x - √xy
3x + y

3x + √xy
9x - y

3x + √xy
3x - y

x
3x - y

those are supposed to be fractions didn't know how to do that

$\frac {\sqrt{x}}{3\sqrt{x} - sqrt{y}}$

Multiply top and bottom by $3\sqrt{x}+sqrt{y}$ . The reason I picked that is $a^2-b^2=(a-b)(a+b)$ . I'm trying to get roots out of the denominator.

$\left( \frac {\sqrt{x}}{3\sqrt{x} - sqrt{y}} \right) \left(\frac {3\sqrt{x}+\sqrt{y}}{3\sqrt{x}+\sqrt{y}}\right) = \frac {3x+\sqrt {xy}}{9x-y}$

I think this is one of your choices.