Answer:
[tex] \longmapsto4 \sqrt{3} + 2 \sqrt{6} .[/tex]
Step-by-step explanation:
[tex]\sf{\dfrac{4\sqrt{3}}{2 - \sqrt{2}}}[/tex]
By Rationalizing the denominator:-
[tex] = \sf{\dfrac{4\sqrt{3}}{2 - \sqrt{2}} \times \dfrac{2 + \sqrt{2}}{2 + \sqrt{2}}}[/tex]
[tex] = \sf{\dfrac{4\sqrt{3}(2 + \sqrt{2})}{(2)^2 - (\sqrt{2})^2}}[/tex]
[tex] = \sf{\dfrac{4\sqrt{3}(2 + \sqrt{2})}{4 - 2}}[/tex]
[tex] = \sf{\dfrac{4\sqrt{3}(2 + \sqrt{2})}{2}}[/tex]
[tex] = \sf{\dfrac{\not{4}\sqrt{3}(2 + \sqrt{2})}{\not{2}}}[/tex]
[tex] = \sf{2\sqrt{3}(2 + \sqrt{2})}[/tex]
[tex] = \sf{4\sqrt{3} + 2\sqrt{6}}[/tex]
[tex] \therefore \sf{\dfrac{4\sqrt{3}}{2 - \sqrt{2}} = 4\sqrt{3} + 2\sqrt{6}}[/tex]
