what is the simplest form of $\frac{2sqrt{2}}{sqrt{3}-sqrt{2}}$?\n$2sqrt{6}+4$\n$2sqrt{5}+4$\n$\frac{2sqrt{6}…

Explanation:

Step1: Rationalize the denominator

Multiply the fraction $\frac{2\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ by $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$. We get $\frac{2\sqrt{2}(\sqrt{3}+\sqrt{2})}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$.

Step2: Expand the numerator and denominator

Expand the numerator: $2\sqrt{2}(\sqrt{3}+\sqrt{2})=2\sqrt{6} + 4$. Expand the denominator using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, so $(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=(\sqrt{3})^{2}-(\sqrt{2})^{2}=3 - 2=1$.

Answer:

$2\sqrt{6}+4$