Answer:
[tex]\huge\boxed{\dfrac{\sqrt6}{\sqrt3-\sqrt2}=3\sqrt2+2\sqrt3}[/tex]
Step-by-step explanation:
[tex]\dfrac{\sqrt6}{\sqrt3-\sqrt2}\\\\\text{use}\ (a-b)(a+b)=a^2-b^2\\\\\dfrac{\sqrt6}{\sqrt3-\sqrt2}\cdot\dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}=\dfrac{\sqrt6(\sqrt3+\sqrt2)}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}=\dfrac{(\sqrt6)(\sqrt3)+(\sqrt6)(\sqrt2)}{(\sqrt3)^2-(\sqrt2)^2}[/tex]
[tex]\text{use}\ \sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\ \text{and}\ (\sqrt{a})^2=a[/tex]
[tex]=\dfrac{\sqrt{(6)(3)}+\sqrt{(6)(2)}}{3-2}=\dfrac{\sqrt{18}+\sqrt{12}}{1}=\sqrt{9\cdot2}+\sqrt{4\cdot3}\\\\=\sqrt9\cdot\sqrt2+\sqrt4\cdot\sqrt3=3\sqrt2+2\sqrt3[/tex]
