Hello, madelynnnnnn!
The question is: \(\displaystyle \L\,\frac{\sqrt{4}}{\sqrt{27 }}\)
then i got: \(\displaystyle \L\,\frac{\sqrt{108}}{27}\;\) . . . right!
then what?
What you did is absolutely correct!
Now see if we can simplify \(\displaystyle \,\sqrt{108}\)
\(\displaystyle \;\;\sqrt{108}\;=\;\sqrt{36\cdot3}\;=\;\sqrt{36}\cdot\sqrt{3}\;=\;6\sqrt{3}\)
So we have: \(\displaystyle \L\,\frac{6\sqrt{3}}{27}\;=\;\frac{2\sqrt{3}}{9}\)
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It would be easier (I think) if you simplified first ... as Denis suggested.
You know that: \(\displaystyle \,\sqrt{4}\,=\,2\)
\(\displaystyle \;\;\)and that: \(\displaystyle \,\sqrt{27}\,=\,\sqrt{9\cdot3}\,=\,\sqrt{9}\cdot\sqrt{3}\,=\,3\sqrt{3}\)
So the problem becomes: \(\displaystyle \L\,\frac{2}{3\sqrt{3}}\)
Now rationalize: \(\displaystyle \L\,\frac{2}{3\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}\:=\:\frac{2\sqrt{3}}{9}\)
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Always simplify first . . . things usually work out much simpler.
Example: \(\displaystyle \L\,\frac{\sqrt{72}}{\sqrt{50}}\)
Instead of messing around with \(\displaystyle 50\) and \(\displaystyle 3600\),
simplify!
\(\displaystyle \;\;\sqrt{72}\:=\:\sqrt{36\cdot2}\:=\:\sqrt{6}\sqrt{2}\:=\:6\sqrt{2}\)
\(\displaystyle \;\;\sqrt{50}\:=\:\sqrt{25\cdot2}\:=\:\sqrt{25}\cdot\sqrt{2}\:=\:5\sqrt{2}\)
The problem becomes: \(\displaystyle \L\;\frac{6\sqrt{2}}{5\sqrt{2}}\;=\;\frac{6}{5}\;\;\) . . . see?