Question Number 192387 by cortano12 last updated on 16/May/23
$$\:\mathrm{Simplify}\: \\ $$$$\:\sqrt{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{4}+\left(\frac{\mathrm{2017}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2017}^{\mathrm{2}} }\right)^{\mathrm{2}} }\right)}\: \\ $$$$\:\mathrm{is}\:….\: \\ $$
Answered by a.lgnaoui last updated on 16/May/23
![(√(4+(((2017^4 −1)/(2017^2 )))^2 ))=(√(4+(2017^2 −(1/(2017^2 )))^2 )) =(√(4+(2017−(1/(2017)))^2 (2017+(1/(2017)))^2 )) =(√(4+[(2017^2 +(1/(2017^2 ))−2)(2017^2 +(1/(217^2 ))+2)]))= =(√((2017^2 +(1/(2017^2 )))^2 )) =2017^2 +(1/(2017^2 )) ⇒(√(2(1+(4+(√((2017^4 −1)/(2017^2 ))))^2 ))= (√(2(1+2017^2 +(1/(2017^2 ))))) =(√(2+2[(2017+(1/(2017)))^2 −2])) =[(√(1+(2017+(1/(2017)))^2 −1)) ](√2) =(2017+(1/(2017)))(√2)](https://www.tinkutara.com/question/Q192391.png)
$$\sqrt{\mathrm{4}+\left(\frac{\mathrm{2017}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2017}^{\mathrm{2}} }\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}+\left(\mathrm{2017}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2017}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{4}+\left(\mathrm{2017}−\frac{\mathrm{1}}{\mathrm{2017}}\right)^{\mathrm{2}} \left(\mathrm{2017}+\frac{\mathrm{1}}{\mathrm{2017}}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{4}+\left[\left(\mathrm{2017}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2017}^{\mathrm{2}} }−\mathrm{2}\right)\left(\mathrm{2017}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{217}^{\mathrm{2}} }+\mathrm{2}\right)\right]}= \\ $$$$=\sqrt{\left(\mathrm{2017}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2017}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:\:\:=\mathrm{2017}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2017}^{\mathrm{2}} } \\ $$$$\Rightarrow\sqrt{\mathrm{2}\left(\mathrm{1}+\left(\mathrm{4}+\sqrt{\frac{\mathrm{2017}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2017}^{\mathrm{2}} }}\right)^{\mathrm{2}} \:\right.}= \\ $$$$\sqrt{\mathrm{2}\left(\mathrm{1}+\mathrm{2017}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2017}^{\mathrm{2}} }\right)}\:=\sqrt{\mathrm{2}+\mathrm{2}\left[\left(\mathrm{2017}+\frac{\mathrm{1}}{\mathrm{2017}}\right)^{\mathrm{2}} −\mathrm{2}\right]} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\sqrt{\mathrm{1}+\left(\mathrm{2017}+\frac{\mathrm{1}}{\mathrm{2017}}\right)^{\mathrm{2}} −\mathrm{1}}\:\right]\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{2017}+\frac{\mathrm{1}}{\mathrm{2017}}\right)\sqrt{\mathrm{2}}\: \\ $$