Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 181447 by Socracious last updated on 25/Nov/22

f(x+(1/x))=x^5 +(1/x^5 ) f(3)=?

$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)=\boldsymbol{\mathrm{x}}^{\mathrm{5}} +\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{5}} } \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{f}}\left(\mathrm{3}\right)=? \\ $$

Answered by Frix last updated on 25/Nov/22

y=x+(1/x) ⇒ x=((y±(√(y^2 −4)))/2)∧x^(−1) =((y∓(√(y^2 −4)))/2) ⇒ f(y)=y^5 −5y^3 +5 f(3)=123

$${y}={x}+\frac{\mathrm{1}}{{x}}\:\Rightarrow \\ $$$${x}=\frac{{y}\pm\sqrt{{y}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\wedge{x}^{−\mathrm{1}} =\frac{{y}\mp\sqrt{{y}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\:\Rightarrow \\ $$$${f}\left({y}\right)={y}^{\mathrm{5}} −\mathrm{5}{y}^{\mathrm{3}} +\mathrm{5} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{123} \\ $$

Answered by manxsol last updated on 25/Nov/22

Answered by Rasheed.Sindhi last updated on 25/Nov/22

x+(1/x)=y x^2 +(1/x^2 )=y^2 −2 x^3 +(1/x^3 )=y^3 −3y (x^2 +(1/x^2 ))(x^3 +(1/x^3 ))=(y^2 −2)(y^3 −3y) x^5 +(1/x^5 )+(1/x)+x=(y^2 −2)(y^3 −3y) x^5 +(1/x^5 )=(y^2 −2)(y^3 −3y)−y f(x)=(x^2 −2)(x^3 −3x)−x f(3)=(9−2)(27−9)−3 =7(18)−3=123

$${x}+\frac{\mathrm{1}}{{x}}={y} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }={y}^{\mathrm{2}} −\mathrm{2} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }={y}^{\mathrm{3}} −\mathrm{3}{y} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)=\left({y}^{\mathrm{2}} −\mathrm{2}\right)\left({y}^{\mathrm{3}} −\mathrm{3}{y}\right) \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\frac{\mathrm{1}}{{x}}+{x}=\left({y}^{\mathrm{2}} −\mathrm{2}\right)\left({y}^{\mathrm{3}} −\mathrm{3}{y}\right) \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\left({y}^{\mathrm{2}} −\mathrm{2}\right)\left({y}^{\mathrm{3}} −\mathrm{3}{y}\right)−{y} \\ $$$${f}\left({x}\right)=\left({x}^{\mathrm{2}} −\mathrm{2}\right)\left({x}^{\mathrm{3}} −\mathrm{3}{x}\right)−{x} \\ $$$${f}\left(\mathrm{3}\right)=\left(\mathrm{9}−\mathrm{2}\right)\left(\mathrm{27}−\mathrm{9}\right)−\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{7}\left(\mathrm{18}\right)−\mathrm{3}=\mathrm{123} \\ $$

Answered by Rasheed.Sindhi last updated on 25/Nov/22

f(x+(1/x))=x^5 +(1/x^5 ) ; f(3)=? x+(1/x)=3 ; x^5 +(1/x^5 )=? (x+(1/x))^2 =3^2 =9 x^2 +(1/x^2 )=9−2=7 (x+(1/x))^3 =3^3 =27 x^3 +(1/x^3 )=27−3(3)=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 )+3=126 x^5 +(1/x^5 )=126−3=123 f(x+(1/x))=x^5 +(1/x^5 ) f(3)=123

$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}\right)=\boldsymbol{\mathrm{x}}^{\mathrm{5}} +\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{5}} }\:;\:\:\boldsymbol{\mathrm{f}}\left(\mathrm{3}\right)=? \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:;\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=? \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} =\mathrm{9} \\ $$$$\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{9}−\mathrm{2}=\mathrm{7} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} =\mathrm{3}^{\mathrm{3}} =\mathrm{27} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{27}−\mathrm{3}\left(\mathrm{3}\right)=\mathrm{18} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)=\left(\mathrm{7}\right)\left(\mathrm{18}\right) \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+{x}+\frac{\mathrm{1}}{{x}}=\mathrm{126} \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }+\mathrm{3}=\mathrm{126} \\ $$$${x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=\mathrm{126}−\mathrm{3}=\mathrm{123} \\ $$$${f}\left({x}+\frac{\mathrm{1}}{{x}}\right)={x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} } \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{123} \\ $$

Commented by manxsol last updated on 25/Nov/22

first :grand idea (x^3 +(1/x^3 ))(x^2 +(1/x^2 ))=(x^5 +(1/x^5 ))+(x+(1/x)) second:desarrollo

$${first}\::{grand}\:{idea}\:\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)+\left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${second}:{desarrollo} \\ $$

Commented by Rasheed.Sindhi last updated on 26/Nov/22

What is meant by ′desarrollo′ sir?

$${What}\:{is}\:{meant}\:{by}\:'{desarrollo}'\:{sir}? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com