[tex]\displaystyle\\
\text{Se da polinomul: }~~~~~f = x^3-5x+1\\
\text{Solutiile ecuatiei:}~~~~~x^3-5x+1=0~~~~~\text{sunt: }~~ x_1,~~x_2,~~x_3 \\\\
\text{Inversele solutiilor sunt: }~~ \frac{1}{x_1},~~\frac{1}{x_2},~~\frac{1}{x_3}\\\\
\text{Suma inverselor solutiilor este: }\\\\
\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3} = ~~\text{(Aducem fractiile la acelasi numitor)}\\\\
=\frac{x_2 x_3}{x_1x_2 x_3}+\frac{x_1x_3}{x_1x_2 x_3}+\frac{x_1x_2}{x_1x_2 x_3}=\frac{x_2 x_3+ x_1x_3+x_1x_2}{x_1x_2 x_3}\\\\
[/tex]
[tex]\displaystyle\\
\text{Folosim Relatiile lui Viete pentru polinomul de gradul 3: }\\\\
\text{Daca avem polinomul:} ~~f=ax^3+bx^2+cx+d\\\\
\text{Relatiile lui Viete sunt:}\\\\
x_1+x_2 +x_3= \frac{-b}{a}\\\\
x_1x_2+x_2 x_3+ x_1x_3=\frac{c}{a}\\\\
x_1x_2 x_3= \frac{-d}{a}\\\\
\text{Polinomul nostru este: }~~f=x^3-5x+1~~\text{in care:}\\
a=1\\
b=0\\
c=-5\\
d=1\\\\
\frac{x_2 x_3+ x_1x_3+x_1x_2}{x_1x_2x_3}=\frac{\dfrac{c}{a}}{\dfrac{-d}{a}}=\frac{\dfrac{-5}{1}}{\dfrac{-1}{1}}= \frac{-5}{-1} =\boxed{\bf 5}
[/tex]
