Question

Buna. Ma poate ajuta cineva sa continui rezolvarea la subpunctul c? Multumesc anticipat!

Answer 1

Te-ai complicat mult, exercițiul ar trebui rezolvat asa:

$\\ f = X^3+5X^2-4$

$x_1+x_2+x_3 = -5 \Rightarrow \begin{cases}x_2+x_3 = -5-x_1\big|:x_1\\x_3+x_1 = -5-x_2\big|:x_2\\x_1+x_2 = -5-x_3\big|:x_3 \end{cases} \Rightarrow$

$\Rightarrow \left.\begin{cases}\frac{x_2+x_3}{x_1} = -\frac{5}{x_1}-1\\\frac{x_3+x_1}{x_2} = -\frac{5}{x_2}-1\\\frac{x_1+x_2}{x_3} = -\frac{5}{x_3}-1 \end{cases}\right|\Rightarrow$

$\displaystyle \Rightarrow \frac{x_2+x_3}{x_1}+\frac{x_3+x_1}{x_2}+\frac{x_1+x_2}{x_3}= -\dfrac{5}{x_1}-\dfrac{5}{x_2}-\dfrac{5}{x_3}-1\cdot 3=$

$=-5\cdot \left(\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{1}{x_3}\right)-3=-5\cdot \dfrac{x_2x_3+x_1x_3+x_1x_2}{x_1x_2x_3}-3 =$

$=-5\cdot \dfrac{0}{4}-3 = \boxed{-3}$