Rationalizing the Denominator

The denominator is $\sqrt{3}$, so we multiply numerator and denominator by $\frac{\sqrt{3}}{\sqrt{3}}$.

$\frac{6}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{6\sqrt{3}}{3}$

$\frac{6\sqrt{3}}{3}=2\sqrt{3}$

The rationalized form is $2\sqrt{3}$.

$6\sqrt{3}$ would only be correct if we multiplied by $\sqrt{3}$ without dividing — this changes the value of the fraction, so it is wrong.

$\frac{6\sqrt{3}}{9}$ would arise from squaring 3 in the denominator (getting 9 instead of 3), which is incorrect.

$3\sqrt{2}$ contains a different radical ($\sqrt{2}$) which cannot arise from this expression.

The only correctly rationalized and simplified form is $2\sqrt{3}$.

$\frac{35}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}=\frac{35\sqrt{7}}{7}$

$\frac{35\sqrt{7}}{7}=5\sqrt{7}$

The simplified rationalized form is $5\sqrt{7}$.

This forgets to divide by 7 after multiplying — it changes the value of the expression.

This incorrectly uses 49 (which is $7^2$) in the denominator instead of 7. $\sqrt{7}\times \sqrt{7}=7$, not 49.

This contains $\sqrt{5}$, which does not appear in the original expression at all.

$5\sqrt{7}$ is the correctly rationalized and simplified form.