it seems that you have this math problem:
\(\displaystyle \L \;\sqrt{18}\,-\,\frac{\sqrt{3}}{\sqrt{6}}\)
Before we get like denominators, let's break this down some.
When doing problems like these, you should first simplify.
The square root of 18 is \(\displaystyle \sqrt{9}\cdot\sqrt{2}=3\sqrt{2}\)
almost \(\displaystyle 3\sqrt{3}\).
So now we have: \(\displaystyle 3\sqrt{2}\,-\,\frac{\sqrt{3}}{\sqrt{6}}\)
We may also simplify the right fraction like this:
\(\displaystyle \frac{\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{3}}{\sqrt{3}\cdot\sqrt{2}}\)
The \(\displaystyle \sqrt{3}\)s cancel, giving us: \(\displaystyle \frac{1}{\sqrt{2}\)
Now we have: \(\displaystyle 3\sqrt{2}\,-\,\frac{1}{\sqrt{2}}\)
We need the like denominator \(\displaystyle \sqrt{2}\).
So just multiply the \(\displaystyle 3\sqrt{2}\) by \(\displaystyle \frac{\sqrt{2}}{\sqrt{2}}\), simplify, and then rationalize the denominator.
