Question Number 220607 by fantastic last updated on 16/May/25
$${symplify} \\ $$$$\frac{\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}+}\sqrt{\mathrm{9}}+\sqrt{\mathrm{15}}}{\:\sqrt{\mathrm{1}}+\sqrt{\mathrm{5}}+\sqrt{\mathrm{12}}} \\ $$
Commented by cadmon98 last updated on 16/May/25
$$ \\ $$is it √5+√9 or √(5+√9)?
Answered by A5T last updated on 16/May/25
$$\frac{\mathrm{3}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}{\mathrm{1}+\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)+\sqrt{\mathrm{5}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}{\mathrm{1}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}} \\ $$$$=\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\right)}{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)+\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}}\right)}=\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{1}}{\frac{\mathrm{a}+\mathrm{b}}{\mathrm{ab}}}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}} \\ $$$$=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+\mathrm{1}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}+\sqrt{\mathrm{3}}}}=\frac{\mathrm{1}}{\frac{\sqrt{\mathrm{3}}−\mathrm{1}+\sqrt{\mathrm{5}}−\sqrt{\mathrm{3}}}{\mathrm{2}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}−\mathrm{1}} \\ $$$$=\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}+\mathrm{1}\right)}{\mathrm{4}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Commented by fantastic last updated on 16/May/25
$${sorry}\:{i}\:{typed}\:{wrong}\:{it}\:{is}\:\sqrt{\mathrm{5}}+\sqrt{\mathrm{9}} \\ $$