Question 326866
1) False, See Answer 234051. 
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2) False, to rationalize you need to yes multiply by 1 but the numerator and denominator have to equal {{{(3-sqrt(3))}}} since the denominator is {{{(3+sqrt(3))}}} so that when you multiply it out the square root terms cancel.
{{{(3-sqrt(3))(3+sqrt(3))=9+3sqrt(3)-3sqrt(3)-3=6}}}

If you only multiply by {{{sqrt(3)}}, then
{{{sqrt(3)(3+sqrt(3))=3sqrt(3)+3}}}
You still have a square root in the denominator.