Question Number 191030 by cortano12 last updated on 16/Apr/23
Answered by Frix last updated on 16/Apr/23
$${t}+\frac{\mathrm{1}}{{t}}=\mathrm{3} \\ $$$$\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} =\mathrm{9}\:\Leftrightarrow\:{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }=\mathrm{7} \\ $$$$\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{3}} =\mathrm{27}\:\Leftrightarrow\:{t}^{\mathrm{3}} +\frac{\mathrm{1}}{{t}^{\mathrm{3}} }=\mathrm{18} \\ $$$${E}=\sqrt{\frac{\left({t}^{\mathrm{3}} +{t}^{\mathrm{2}} \right)\left({t}^{\mathrm{5}} +\mathrm{1}\right)}{{t}^{\mathrm{5}} }}=\sqrt{\frac{{t}^{\mathrm{8}} +{t}^{\mathrm{7}} +{t}^{\mathrm{3}} +{t}^{\mathrm{2}} }{{t}^{\mathrm{5}} }}= \\ $$$$=\sqrt{{t}^{\mathrm{3}} +\frac{\mathrm{1}}{{t}^{\mathrm{3}} }+{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}=\sqrt{\mathrm{18}+\mathrm{7}}=\sqrt{\mathrm{25}}=\mathrm{5} \\ $$