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f-x-1-x-x-5-1-x-5-f-3-Q-181447-reposted-for-a-different-answer-

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Question Number 181524 by Rasheed.Sindhi last updated on 26/Nov/22
f(x+(1/x))=x^5 +(1/x^5 ) ; f(3)=? Q#181447 reposted for a different answer.
f(x+1x)=x5+1x5;f(3)=?You can't use 'macro parameter character #' in math mode
Answered by Rasheed.Sindhi last updated on 26/Nov/22
f(x+(1/x))=x^5 +(1/x^5 ) ; f(3)=? x+(1/x)=3⇒ { ((x=3−(1/x))),(((1/x)=3−x)) :} x^5 =? x=3−(1/x) ⇒x^2 =3x−1 ⇒x^3 =3x^2 −x=3(3x−1)−x=8x−3 ⇒x^4 =8x^2 −3x=8(3x−1)−3x=21x−8 ⇒x^5 =21x^2 −8x=21(3x−1)−8x=55x−21 (1/x^5 )=? (1/x)=3−x (1/x^2 )=((3−x)/x)=3((1/x))−1=3(3−x)−1=8−3x (1/x^3 )=((8−3x)/x)=8((1/x))−3=8(3−x)−3=21−8x (1/x^4 )=21((1/x))−8=21(3−x)−8=55−21x (1/x^5 )=55((1/x))−21=55(3−x)−21=144−55x x^5 +(1/x^5 )=(55x−21)+(144−55x)=123 f(3)=123
f(x+1x)=x5+1x5;f(3)=?x+1x=3⇒{x=3−1x1x=3−xx5=?x=3−1x⇒x2=3x−1⇒x3=3x2−x=3(3x−1)−x=8x−3⇒x4=8x2−3x=8(3x−1)−3x=21x−8⇒x5=21x2−8x=21(3x−1)−8x=55x−211x5=?1x=3−x1x2=3−xx=3(1x)−1=3(3−x)−1=8−3x1x3=8−3xx=8(1x)−3=8(3−x)−3=21−8x1x4=21(1x)−8=21(3−x)−8=55−21x1x5=55(1x)−21=55(3−x)−21=144−55xx5+1x5=(55x−21)+(144−55x)=123f(3)=123
Answered by SEKRET last updated on 26/Nov/22
x+(1/x)=3 → x^5 +(1/x^5 )=? x^2 +(1/x^2 )=7 x^3 +(1/x^3 )+3∙(x+(1/x))=27 x^3 +(1/x^3 )=27−9=18 (x^2 +(1/x^2 ))(x^3 +(1/x^3 ))=7∙18 x^5 +(1/x^5 )+x+(1/x)=126 x^5 +(1/x^5 )=123 ABDULAZIZ ABDUVALIYEV
\boldsymbolx+1\boldsymbolx=3→\boldsymbolx5+1\boldsymbolx5=?\boldsymbolx2+1\boldsymbolx2=7\boldsymbolx3+1\boldsymbolx3+3⋅(\boldsymbolx+1\boldsymbolx)=27\boldsymbolx3+1\boldsymbolx3=27−9=18(\boldsymbolx2+1\boldsymbolx2)(\boldsymbolx3+1\boldsymbolx3)=7⋅18\boldsymbolx5+1\boldsymbolx5+\boldsymbolx+1\boldsymbolx=126\boldsymbolx5+1\boldsymbolx5=123\boldsymbolABDULAZIZ\boldsymbolABDUVALIYEV

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