Question Number 146964 by mathdanisur last updated on 16/Jul/21 | ||
$${Simplify}: \\ $$$$\frac{\sqrt{\mathrm{2}}\:\centerdot\:\sqrt{\mathrm{2}\:+\:\sqrt{\mathrm{2}}}\:\centerdot\:\sqrt{\mathrm{2}\:-\:\sqrt{\mathrm{2}}}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}}\:=\:? \\ $$ | ||
Answered by Olaf_Thorendsen last updated on 16/Jul/21 | ||
$${x}\:=\:\frac{\sqrt{\mathrm{2}}\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}.\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}} \\ $$$${x}\:=\:\frac{\sqrt{\mathrm{2}}\sqrt{\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}} \\ $$$${x}\:=\:\frac{\sqrt{\mathrm{2}}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}\sqrt{\mathrm{2}}}\:=\:\sqrt{\sqrt{\mathrm{2}}}\:=\:\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$ | ||
Commented by mathdanisur last updated on 16/Jul/21 | ||
$${thank}\:{you}\:{Ser} \\ $$ | ||