Question

Daca $x+ \frac{1}{x} = 2$ , calculati $x$ + $x^{2} + x^{3} + x^{4} + \frac{1}{x} + \frac{1}{ x^{2} } + \frac{1}{ x^{3} } + \frac{1}{ x^{4} }$

Answer 1

$x+ \frac{1}{x}=2 \\ \frac{x^2-2x+1}{x}=0 \\ x=1$
[tex]x+ x^{2} +x^3+x^4+ \frac{1}{x}+ \frac{1}{ x^{2} } + \frac{1}{x^3}+ \frac{1}{x^4}=8 [/tex]

Answer 2

= ( x + 1 / x)  + ( x² + 1 / x² ) + ( x³ + 1 /x³ ) + ( x⁴ + 1 / x⁴ )  = 2 · 4 = 8 
       =2                 4 -2 =2             = 2               = 2 
x³ + 1 /x³ = [ x + 1 /x ] · [ x² - x · 1 /x + 1 /x² ]
             =           2  · [  2 -  1 ] = 2
( x² + 1 / x² )² =  2 
x⁴ + 2· x²· 1 /x²  + 1 / x⁴ = 2² = 4 
          = 2