Question

After rationalizing the numerator of $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}$ the denominator in simplest form is

• A. $\sqrt{3}(\sqrt{3}+\sqrt{2})$
• B. $\sqrt{3}(\sqrt{3}-\sqrt{2})$
• C. $3-\sqrt{3}\sqrt{2}$
• D. $\frac{1}{3+\sqrt{6}}$
• E. None of these

Answer 1

The correct option is D $\frac{1}{3+\sqrt{6}}$

$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}\cdot \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{1}{3+\sqrt{6}}$.