Help please: { ((x+x^(−1) −(√2)=0)),((x^(2017) +x^(2017) +(√2)=?)) :}


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 184424 by Fridunatjan08 last updated on 06/Jan/23
$Help please: { ((x+x^(−1) −(√2)=0)),((x^(2017) +x^(2017) +(√2)=?)) :}$
$H e l p p l e a s e :$ ${\begin{cases} x + x^{- 1} - \sqrt{2} = 0 \\ x^{2017} + x^{2017} + \sqrt{2} = ? \end{cases}$
Commented by Rasheed.Sindhi last updated on 06/Jan/23

$x^{2017} + x^{- 2017} + \sqrt{2} = ? ?$
	Answered by Rasheed.Sindhi last updated on 06/Jan/23
	${ ((x+x^(−1) −(√2)=0)),((x^(2017) +x^(−2017) +(√2)=?)) :} x(x+x^(−1) −(√2) =0)⇒x^2 =(√2) x−1 x^3 =(√2) x^2 −x=(√2) ((√2) x−1)−x=x−(√2) x^4 =x^2 −(√2) x=(√2) x−1−(√2) x=−1 x^(2017) +x^(−2017) +(√2) =(x^4 )^(504) x+(x^4 )^(−504) x^(−1) +(√2) =(−1)^(504) x+(−1)^(−504) x^(−1) +(√2) =x+x^(−1) +(√2) =(√2) +(√2) =2(√(2 ))✓$
	${\begin{cases} x + x^{- 1} - \sqrt{2} = 0 \\ x^{2017} + x^{- 2017} + \sqrt{2} = ? \end{cases}$ $x (x + x^{- 1} - \sqrt{2} = 0) \Rightarrow x^{2} = \sqrt{2} x - 1$ $x^{3} = \sqrt{2} x^{2} - x = \sqrt{2} (\sqrt{2} x - 1) - x = x - \sqrt{2}$ $x^{4} = x^{2} - \sqrt{2} x = \sqrt{2} x - 1 - \sqrt{2} x = - 1$ $x^{2017} + x^{- 2017} + \sqrt{2}$ $= {(x^{4})}^{504} x + {(x^{4})}^{- 504} x^{- 1} + \sqrt{2}$ $= {(- 1)}^{504} x + {(- 1)}^{- 504} x^{- 1} + \sqrt{2}$ $= x + x^{- 1} + \sqrt{2} = \sqrt{2} + \sqrt{2} = 2 \sqrt{2} ✓$
	Answered by mr W last updated on 06/Jan/23

	$x + \frac{1}{x} = \sqrt{2}$ $\Rightarrow x^{2} + \frac{1}{x^{2}} + 2 = 2$ $\Rightarrow x^{2} + \frac{1}{x^{2}} = 0$ $\Rightarrow x^{4} = - 1$ $x^{2017} + \frac{1}{x^{2017}} + \sqrt{2}$ $= {(x^{4})}^{504} x + \frac{1}{{(x^{4})}^{504} x} + \sqrt{2}$ $= x + \frac{1}{x} + \sqrt{2}$ $= \sqrt{2} + \sqrt{2}$ $= 2 \sqrt{2}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum