Answer
$\dfrac{5\sqrt3+\sqrt 6}{23}$
Work Step by Step
Rationalize the denominator by multiplying $5+\sqrt2$ to both the numerator and the denominator:
$\dfrac{\sqrt 3}{5-\sqrt 2}\\
=\dfrac{\sqrt 3}{5-\sqrt 2} \cdot \dfrac{5+\sqrt 2}{5+\sqrt 2}$
Use special formula $(a-b)(a+b)=a^2-b^2$ in the denominator.
$=\dfrac{\sqrt 3(5+\sqrt 2)}{5^2-(\sqrt 2)^2}$
$=\dfrac{\sqrt 3(5+\sqrt 2)}{25-2}$
$=\dfrac{\sqrt 3(5+\sqrt 2)}{23}$
Use the distributive property:
$=\dfrac{5\sqrt3+\sqrt 2\cdot \sqrt3}{23}$
$=\dfrac{5\sqrt3+\sqrt {2(3)}}{23}$
$=\dfrac{5\sqrt3+\sqrt6}{23}$
Hence, the correct answer is $\frac{5\sqrt3+\sqrt 6}{23}$.
