Answer
$6-3\sqrt3$
Work Step by Step
Rationalize the denominator by multiplying the conjugate of the denominator (which is $2-\sqrt3$) to both the numerator and the denominator to obtain:
$=\dfrac{3(2-\sqrt3)}{(2+\sqrt3)(2-\sqrt3)}
\\=\dfrac{6-3\sqrt3}{(2+\sqrt3)(2-\sqrt3)}$
Use the rule $(a-b)(a+b)=a^2-b^2$ to obtain:
$=\dfrac{6-3\sqrt3}{2^2-(\sqrt3)^2}
\\=\dfrac{6-3\sqrt3}{4-3}
\\=\dfrac{6-3\sqrt3}{1}
\\=6-3\sqrt3$
