Answer:
Step-by-step explanation:
[tex]\dfrac{5+2\sqrt3}{7+4\sqrt3}\qquad\text{use}\ (a+b)(a-b)=a^2-b^2\\\\=\dfrac{5+2\sqrt3}{7+4\sqrt3}\cdot\dfrac{7-4\sqrt3}{7-4\sqrt3}=\dfrac{(5+2\sqrt3)(7-4\sqrt3)}{7^2-(4\sqrt3)^2}\qquad\text{use FOIL}\\\\=\dfrac{(5)(7)+(5)(-4\sqrt3)+(2\sqrt3)(7)+(2\sqrt3)(-4\sqrt3)}{49-4^2(\sqrt3)^2}\\\\=\dfrac{35-20\sqrt3+14\sqrt3-8(3)}{49-(16)(3)}=\dfrac{(35-24)+(-20\sqrt3+14\sqrt3)}{49-48}\\\\=\dfrac{11-6\sqrt3}{1}=11-6\sqrt3[/tex]