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Question-191030




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

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