Answer:
[tex]\large\boxed{\dfrac{2\sqrt3}{2+3\sqrt2}=\dfrac{3\sqrt6}{7}-\dfrac{2\sqrt3}{7}}[/tex]
Step-by-step explanation:
[tex]\dfrac{2\sqrt3}{2+3\sqrt2}\\\\\text{use}\ (a+b)(a-b)=a^2-b^2\\\\=\dfrac{2\sqrt3}{2+3\sqrt2}\cdot\dfrac{2-3\sqrt2}{2-3\sqrt2}=\dfrac{2\sqrt3(2-3\sqrt2)}{2^2-(3\sqrt2)^2}\\\\\text{use the distributive property}\ a(b+c)=ab+ac\\\\=\dfrac{(2\sqrt3)(2)+(2\sqrt3)(-3\sqrt2)}{4-3^2(\sqrt2)^2}\\\\\text{use}\ (\sqrt{a})^2=a\ \text{and}\ \sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\\\\=\dfrac{4\sqrt3-6\sqrt6}{4-(9)(2)}=\dfrac{4\sqrt3-6\sqrt6}{4-18}=\dfrac{4\sqrt3-6\sqrt6}{-14}=-\dfrac{4\sqrt3}{14}+\dfrac{6\sqrt6}{14}\\\\=\dfrac{3\sqrt6}{7}-\dfrac{2\sqrt3}{7}[/tex]
