Question

Please help i dont understand what its asking

Answer 1

Multiply both the numerator and denominator by $2+\sqrt3$, which is called the "conjugate" of $2-\sqrt3$:

$\dfrac{5+\sqrt3}{2-\sqrt3}\cdot\dfrac{2+\sqrt3}{2+\sqrt3}$

Why do we pick this number? Recall the difference of squares factorization:

$a^2-b^2=(a-b)(a+b)$

Now replace $a=2$ and $b=\sqrt3$. Then

$(2-\sqrt3)(2+\sqrt3)=2^2-(\sqrt3)^2=4-3=1$

So in your fraction, you end up with

$\dfrac{5+\sqrt3}{2-\sqrt3}=\dfrac{(5+\sqrt3)(2+\sqrt3)}1=(5+\sqrt3)(2+\sqrt3)$

Finally, just expand the product.

$(5+\sqrt3)(2+\sqrt3)=10+7\sqrt3+(\sqrt3)^2=\boxed{13+7\sqrt3}$