Răspuns :
[tex]x+ \frac{1}{x}=2 \\ \frac{x^2-2x+1}{x}=0 \\ x=1 [/tex]
[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]
[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]
= ( 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
=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