Question

Solve the problems :

Answer 1

$\large\mathscr \color{white}\colorbox{red}{\colorbox{orange}{\colorbox{yellow}{\colorbox{green}{\colorbox{blue}{\colorbox{indigo}{\colorbox{violet}{\colorbox{black}{\underline {ANSWER}}}}}}}}}$

19.

$\tt{8 - 2 \sqrt{15 }} \\ \tt = 5 + 3 - 2 \sqrt{15} \\ \tt = {(\sqrt{5} )} ^{2} + {( \sqrt{3)} }^{2} - 2 \times \sqrt{5} \times \sqrt{3} \\ \\ \tt = { (\sqrt{5} - \sqrt{3}) }^{2} \\ \\ \tt{\therefore8 - 2 \sqrt{15} = {( \sqrt{5} - \sqrt{3} ) }^{2} }\\ \\\tt \sqrt{8 - 2\sqrt{15} } = \underline{\underline{ \sqrt{5} - \sqrt{3}}}$

20.

$\tt\frac{1}{1 + \sqrt{2} + \sqrt{3} } \\ \\ = \small \tt\frac{1}{(( 1 + \sqrt{2}) + \sqrt{3} )} \times \frac{((1 + \sqrt{2}) - \sqrt{3} ) }{((1 + \sqrt{2}) - \sqrt{3} )} \\ \\ = \small \tt \frac{1 + \sqrt{2} - \sqrt{3} }{ {(1 + \sqrt{2}) }^{2} - (1 + \sqrt{2} )( \sqrt{3}) + \sqrt{3}(1 + \sqrt{2} ) - { \sqrt{3} }^{2} } \\ \\ = \: \small \tt \frac{1 + \sqrt{2} - \sqrt{3} }{1 + 2 + 2 \sqrt{2} \cancel{\color{red}{- \sqrt{3} }} \cancel{\color{red}{- \sqrt{6} }}+ \cancel{\color{red}{\sqrt{3} }}+ \cancel{\color{red}{\sqrt{6}} } - 9} \\ \\ = \tt \frac{1 + \sqrt{2} - \sqrt{3} }{2 \sqrt{2 } - 6 } \\ \\ = \tt \frac{(1+\sqrt{2}-\sqrt{3})}{(2\sqrt{2}-6)}\times\frac{2\sqrt{2}+6)}{(2\sqrt{2}+6)}\\ \\\tt=\frac{2\sqrt{2}+6+2{(\sqrt{2})}^{2}+6\sqrt{2}-2\sqrt{2}\sqrt{3}-6\sqrt{3}}{({2\sqrt{2}})^{2}-{6}^{2}}\\\\\tt\frac{8\sqrt{2}+10-2\sqrt{2}\sqrt{3}-6\sqrt{3}}{8-36}\\\\=\tt\underline{\underline{\frac{8\sqrt{2}+10-2\sqrt{6}-6\sqrt{3}}{-28}}}$

21.

$\tiny3(\frac{1}{\sqrt{1}+\sqrt{2}}\times\frac{\sqrt{1}-\sqrt{2}}{\sqrt{1}-\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}\times\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}\times\frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}+...+\frac{1}{\sqrt{7}+\sqrt{8}}\times\frac{\sqrt{7}-\sqrt{8}}{\sqrt{7}-\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{9}}\times\frac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}})\\\\3(\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{7}-\sqrt{8}}{-1}+\frac{\sqrt{8}-\sqrt{9}}{-1})$