1. Rationalize the denominator of the following:

Here are the solutions for rationalizing the denominator for each expression: 1. Rationalize the denominator of the following: i) $\frac{1}{2+\sqrt{3}}$ Step 1: Multiply the numerator and denominator by the conjugate of the denominator, which is $2-\sqrt{3}$. $$ \frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} $$ Step 2: Apply the difference of squares formula $(a+b)(a-b) = a^2-b^2$ in the denominator. $$ \frac{2-\sqrt{3}}{(2)^2 - (\sqrt{3})^2} = \frac{2-\sqrt{3}}{4-3} $$ Step 3: Simplify the expression. $$ \frac{2-\sqrt{3}}{1} = 2-\sqrt{3} $$ The rationalized expression is $\boxed{2-\sqrt{3}}$. ii) $\frac{1}{3+2\sqrt{2}}$ Step 1: Multiply the numerator and denominator by the conjugate of the denominator, which is $3-2\sqrt{2}$. $$ \frac{1}{3+2\sqrt{2}} \times \frac{3-2\sqrt{2}}{3-2\sqrt{2}} $$ Step 2: Apply the difference of squares formula in the denominator. $$ \frac{3-2\sqrt{2}}{(3)^2 - (2\sqrt{2})^2} = \frac{3-2\sqrt{2}}{9 - (4 \times 2)} $$ Step 3: Simplify the expression. $$ \frac{3-2\sqrt{2}}{9-8} = \frac{3-2\sqrt{2}}{1} = 3-2\sqrt{2} $$ The rationalized expression is $\boxed{3-2\sqrt{2}}$. iii) $\frac{1}{4\sqrt{3}-5}$ Step 1: Multiply the numerator and denominator by the conjugate of the denominator, which is $4\sqrt{3}+5$. $$ \frac{1}{4\sqrt{3}-5} \times \frac{4\sqrt{3}+5}{4\sqrt{3}+5} $$ Step 2: Apply the difference of squares formula in the denominator. $$ \frac{4\sqrt{3}+5}{(4\sqrt{3})^2 - (5)^2} = \frac{4\sqrt{3}+5}{(16 \times 3) - 25} $$ Step 3: Simplify the expression. $$ \frac{4\sqrt{3}+5}{48-25} = \frac{4\sqrt{3}+5}{23} $$ The rationalized expression is $\boxed{\frac{4\sqrt{3}+5}{23}}$. iv) $\frac{16}{2\sqrt{3}+\sqrt{11}}$ Step 1: Multiply the numerator and denominator by the conjugate of the denominator, which is $2\sqrt{3}-\sqrt{11}$. $$ \frac{16}{2\sqrt{3}+\sqrt{11}} \times \frac{2\sqrt{3}-\sqrt{11}}{2\sqrt{3}-\sqrt{11}} $$ Step 2: Apply the difference of squares formula in the denominator. $$ \frac{16(2\sqrt{3}-\sqrt{11})}{(2\sqrt{3})^2 - (\sqrt{11})^2} = \frac{16(2\sqrt{3}-\sqrt{11})}{(4 \times 3) - 11} $$ Step 3: Simplify the expression. $$ \frac{16(2\sqrt{3}-\sqrt{11})}{12-11} = \frac{16(2\sqrt{3}-\sqrt{11})}{1} = 16(2\sqrt{3}-\sqrt{11}) $$ $$ = 32\sqrt{3}-16\sqrt{11} $$ The rationalized expression is $\boxed{32\sqrt{3}-16\sqrt{11}}$. v) $\frac{9-\sqrt{2}}{9+\sqrt{2}}$ Step 1: Multiply the numerator and denominator by the conjugate of the denominator, which is $9-\sqrt{2}$. $$ \frac{9-\sqrt{2}}{9+\sqrt{2}} \times \frac{9-\sqrt{2}}{9-\sqrt{2}} $$ Step 2: Expand the numerator using $(a-b)^2 = a^2-2ab+b^2$ and the denominator using $(a+b)(a-b) = a^2-b^2$. $$ \frac{(9-\sqrt{2})^2}{(9)^2 - (\sqrt{2})^2} = \frac{9^2 - 2(9)(\sqrt{2}) + (\sqrt{2})^2}{81 - 2} $$ Step 3: Simplify the expression. $$ \frac{81 - 18\sqrt{2} + 2}{79} = \frac{83 - 18\sqrt{2}}{79} $$ The rationalized expression is $\boxed{\frac{83 - 18\sqrt{2}}{79}}$. vi) $\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}}$ Step 1: Multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{13}+\sqrt{11}$. $$ \frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}-\sqrt{11}} \times \frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}+\sqrt{11}} $$ Step 2: Expand the numerator using $(a+b)^2 = a^2+2ab+b^2$ and the denominator using $(a-b)(a+b) = a^2-b^2$. $$ \frac{(\sqrt{13}+\sqrt{11})^2}{(\sqrt{13})^2 - (\sqrt{11})^2} = \frac{(\sqrt{13})^2 + 2(\sqrt{13})(\sqrt{11}) + (\sqrt{11})^2}{13 - 11} $$ Step 3: Simplify the expression. $$ \frac{13 + 2\sqrt{143} + 11}{2} = \frac{24 + 2\sqrt{143}}{2} $$ Step 4: Factor out 2 from the numerator and simplify. $$ \frac{2(12 + \sqrt{143})}{2} = 12 + \sqrt{143} $$ The rationalized expression is $\boxed{12 + \sqrt{143}}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.