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

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

Answered by Frix last updated on 25/Nov/22

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

Answered by manxsol last updated on 25/Nov/22

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

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

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

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

Commented by manxsol last updated on 25/Nov/22

$$first:grandidea(x^3+\frac{1}{x^3})(x^2+\frac{1}{x^2})=(x^5+\frac{1}{x^5})+(x+\frac{1}{x})$$$$second:desarrollo$$

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

$$Whatismeantby{}^{\prime }{desarrollo}^{\prime }sir?$$

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