Question

Alguém me ajuda? Assunto: Relação de Girard.

Answer 1

Bom dia!

Solução!

Observe que a relação de Girard já esta colocada,agora precisa encaixar o polinômio na relação, para isso precisamos conhecer seus coeficientes.
Vou chamar alfa,beta e gama de a,b e c.

$3 x^{3}-13 x^{2} +11x-11=0\\\\\\\ a=3\\\\\\\ b=-13\\\\\\\ c=11\\\\\\\ d=-8\\\\\\\\\\\\\\ a+b+c= \frac{13}{3}\\\\\\ ab+ac+bc= \frac{11}{3}\\\\\\\ a.b.c= \frac{8}{3}$

Veja a expressão! Para chegarmos ao resultado,nas duas partes do lado direito vamos ter que simplificar.Vejamos como.

$y=(a+b+c)+(ab+bc+ac)+\bigg( \dfrac{1}{a}+ \dfrac{1}{b}+ \dfrac{1}{c}\bigg)+\bigg( \dfrac{1}{ab}+ \dfrac{1}{bc}+ \dfrac{1}{ac}\bigg)\\\\\\\\\\\\\ \boxed{\bigg( \dfrac{1}{a}+ \dfrac{1}{b}+ \dfrac{1}{c}\bigg)=\bigg( \dfrac{bc+ac+ab}{a.b.c}\bigg)}\\\\\\\\\\\\ \boxed{\bigg( \dfrac{1}{ab}+ \dfrac{1}{bc}+ \dfrac{1}{ac}\bigg)=(a+b+c). \bigg(\dfrac{1}{a}+ \dfrac{1}{b}+ \dfrac{1}{c}\bigg)=\bigg(\dfrac{a+b+c}{a.b.c} \bigg)}$

Reescrevendo a expressão fica assim!

$y=(a+b+c)+(ab+bc+ac)+\bigg( \dfrac{bc+ac+ab}{a.b.c}\bigg)+ \bigg(\dfrac{a+b+c}{a.b.c}\bigg)\\\\\\\\\ y= \dfrac{13}{3}+ \dfrac{11}{3}+ \dfrac{11}{ \dfrac{3}{ \dfrac{8}{3} } }+ \dfrac{13}{ \dfrac{3}{ \dfrac{8}{3} } }\\\\\\\\\ y= \dfrac{13}{3}+ \dfrac{11}{3}+ \dfrac{11}{3}. \frac{3}{8}+ \frac{13}{3}. \frac{3}{8}\\\\\\\\ y= \dfrac{13}{3}+ \dfrac{11}{3}+ \dfrac{11}{8}+ \frac{13}{8}\\\\\\\\ y= \dfrac{104+88+33+39}{24}\\\\\\\ y= \dfrac{264}{24} \\\\\\ y=11$


$\boxed{Resposta~~y=11~~\boxed{Alternativa~~~d}}$


Bom dia!
Bons estudos!