Question

Ex ăsta vă rog….. vreau rezolvare completă…….

Answer 1

$\it x^3+x+m=0 \Rightarrow x^3=-x-m\ \ \ \ (1)\\ \\ x^3=-x-m\Big|_{\cdot x^2} \Rightarrow x^5=-x^3-mx^2 \stackrel{(1)}{\Longrightarrow}x^5=-(-x-m)-mx^2 \Rightarrow \\ \\ \\ \Rightarrow x^5=-mx^2+x+m\ \ \ \ \ (2)\\ \\ \\ (2) \Rightarrow x_1^5+x_2^5++x_3^5=-m(x_1^2+x_2^2+x_3^2)+x_1+x_2+x_3+3m\ \ \ \ \ (3)$

Cu formulele lui Viète, avem:

$\it x_1+x_2+x_3=0,\ iar\ rela\c{\it t}ia\ (3)\ devine:\\ \\ x_1^5+x_2^5+x_3^5=-m(x_1^2+x_2^2+x_3^2)+3m\ \ \ \ \ (3')\\ \\ \\ x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3)^2-2(x_1x_2+x_2x_3+x_3x_1)\ \stackrel{Vi\grave{e}te}{\Longrightarrow}\\ \\ \Rightarrow x_1^2+x_2^2+x_3^2=-2\cdot1=-2,\ ,\ iar\ rela\c{\it t}ia\ (3')\ devine:\\ \\ \\x_1^5+x_2^5+x_3^5=2m+3m=5m\ \ \ (3'')$

Relația din enunțul problemei devine: 5m = 10 ⇒ m=2 .

Răspunsul corect este D.  2