Question

$\frac{1}{3+\sqrt{10}}+\frac{1}{\sqrt{10}+\sqrt{11}}+\frac{1}{\sqrt{11}+2\sqrt{3}}+\frac{1}{2\sqrt{3}+\sqrt{13}}+\frac{1}{\sqrt{13}+\sqrt{14}}+\frac{1}{\sqrt{14}+\sqrt{15}}+\frac{1}{\sqrt{15}+4}$

Answer 1

Dear student
$$\begin{gathered} \frac{1}{3+\sqrt{10}}+\frac{1}{\sqrt{10}{\displaystyle +}{\displaystyle \sqrt{11}}}+\frac{1}{\sqrt{11}{\displaystyle +}{\displaystyle 2}{\displaystyle \sqrt{3}}}+\frac{1}{2\sqrt{3}+\sqrt{13}}+\frac{1}{\sqrt{13}{\displaystyle +}{\displaystyle \sqrt{14}}}+\frac{1}{\sqrt{14}{\displaystyle +}{\displaystyle \sqrt{15}}}+\frac{1}{\sqrt{15}{\displaystyle +}{\displaystyle 4}} \\ =\frac{{\displaystyle 1}}{{\displaystyle 3+\sqrt{10}}}\times \frac{3-\sqrt{10}}{3-\sqrt{10}}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{10}+\sqrt{11}}}\times \frac{\sqrt{10}{\displaystyle -}\sqrt{11}}{\sqrt{10}{\displaystyle -}\sqrt{11}}+\frac{1}{\sqrt{11}{\displaystyle +}{\displaystyle 2}{\displaystyle \sqrt{3}}}\times \frac{\sqrt{11}{\displaystyle -}2\sqrt{3}}{\sqrt{11}{\displaystyle -}2\sqrt{3}} \\ +\frac{1}{2\sqrt{3}+\sqrt{13}}\times \frac{2\sqrt{3}-\sqrt{13}}{2\sqrt{3}-\sqrt{13}}+\frac{1}{\sqrt{13}{\displaystyle +}{\displaystyle \sqrt{14}}}\times \frac{\sqrt{13}{\displaystyle -}\sqrt{14}}{\sqrt{13}{\displaystyle -}\sqrt{14}}+\frac{1}{\sqrt{14}{\displaystyle +}{\displaystyle \sqrt{15}}}\times \frac{\sqrt{14}{\displaystyle -}\sqrt{15}}{\sqrt{14}{\displaystyle -}\sqrt{15}}+\frac{1}{\sqrt{15}{\displaystyle +}{\displaystyle 4}}\times \frac{\sqrt{15}{\displaystyle -}4}{\sqrt{15}{\displaystyle -}4} \\ =\frac{3-\sqrt{10}}{9-10}+\frac{\sqrt{10}-\sqrt{11}}{10-11}+\frac{{\displaystyle \sqrt{11}-2\sqrt{3}}}{{\displaystyle 11-12}}+\frac{{\displaystyle 2\sqrt{3}-\sqrt{13}}}{{\displaystyle 12-13}}+\frac{{\displaystyle \sqrt{13}-\sqrt{14}}}{13{\displaystyle -}14}+\frac{{\displaystyle \sqrt{14}-\sqrt{15}}}{14-15}+\frac{{\displaystyle \sqrt{15}-4}}{15-16} \\ =\cancel{\sqrt{10}}-3+\cancel{\sqrt{11}}-\cancel{\sqrt{10}}+\cancel{2\sqrt{3}}-\cancel{\sqrt{11}}+\cancel{\sqrt{13}}-\cancel{2\sqrt{3}}+\cancel{\sqrt{14}}-\cancel{\sqrt{13}}+\sqrt{\cancel{15}}-\cancel{\sqrt{14}}+4-\cancel{\sqrt{15}} \\ =1 \end{gathered}$$
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