Question

42. Determine os elementos dos conjuntos abaixo: ​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\sf A = \left\{\sf x \in \mathbb{Z}~|~3x^2- 4x - 4 = 0\right \}$

$\sf 3x^2- 4x - 4 = 0$

$\sf 3x^2- 4x + 6x - 6x - 4 = 0$

$\sf 3x^2 + 2x - 6x - 4 = 0$

$\sf x(3x + 2) - 2(3x + 2) = 0$

$\sf (x - 2)\:.\:(3x + 2) = 0$

$\sf x - 2 = 0 \rightarrow x = 2$

$\sf 3x + 2 = 0 \rightarrow x = -\dfrac{2}{3}$

$\boxed{\boxed{\sf A = \left\{2\right\}}}$

$\sf B = \left\{\sf y \in \mathbb{I}~|~y^2 - 7 = 0\right \}$

$\sf y^2 - 7 = 0$

$\sf y^2 = 7$

$\sf y = \pm\:\sqrt{7}$

$\boxed{\boxed{\sf B = \left\{\sqrt{7}\:,\:-\sqrt{7}\right\}}}$

$\sf C = \left\{\sf a \in \mathbb{N}~|~\dfrac{2}{a} + a = 3\right \}$

$\sf \dfrac{2}{a} + a = 3$

$\sf 2 + a^2 = 3a$

$\sf a^2 - 3a + 2 = 0$

$\sf a^2 - 3a + a - a + 2 = 0$

$\sf a^2 - 2a - a + 2 = 0$

$\sf a(a - 2) - 1(a - 2) = 0$

$\sf (a - 1)\:.\:(a - 2) = 0$

$\sf a - 1 = 0 \rightarrow a = 1$

$\sf a - 2 = 0 \rightarrow a = 2$

$\boxed{\boxed{\sf C = \left\{1\:,\:2\right\}}}$

$\sf D = \left\{\sf x \in \mathbb{Q}~|~3 + x^2 = 4\right \}$

$\sf 3 + x^2 = 4$

$\sf x^2 = 1$

$\sf x = \pm\:\sqrt{1}$

$\sf x = \pm\:1$

$\boxed{\boxed{\sf D = \left\{1\:,\:-1\right\}}}$

$\sf E = \left\{\sf y \in \mathbb{Q}~|~y^2 - y - 1 = 0\right \}$

$\sf y^2 - y - 1 = 0$

$\sf a = 1 \Leftrightarrow b = -1 \Leftrightarrow c = -1$

$\sf \Delta = b^2 - 4.a.c$

$\sf \Delta = (-1)^2 - 4.1.(-1)$

$\sf \Delta = 1 + 4$

$\sf \Delta = 5$

$\sf{y = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{1 \pm \sqrt{5}}{2} \rightarrow \begin{cases}\sf{y' = \dfrac{1 + \sqrt{5}}{2}\\\\\sf{y'' = \dfrac{1 - \sqrt{5}}{2}\end{cases}}$

$\boxed{\boxed{\sf E = \left\{\emptyset\right\}}}$

$\sf F = \left\{\sf x \in \mathbb{C}~|~x^2 + 4 = 0\right \}$

$\sf x^2 + 4 = 0$

$\sf x^2 = -4$

$\sf x = \pm\:\sqrt{-4}$

$\sf x = \pm\:2i$

$\boxed{\boxed{\sf F = \left\{2i\:,-2i\right\}}}$

$\sf G = \left\{\sf x \in \mathbb{C}~|~x^2- 6x + 13 = 0\right \}$

$\sf x^2- 6x + 13 = 0$

$\sf a = 1 \Leftrightarrow b = -6 \Leftrightarrow c = 13$

$\sf \Delta = b^2 - 4.a.c$

$\sf \Delta = (-6)^2 - 4.1.13$

$\sf \Delta = 36 - 52$

$\sf \Delta = -16$

$\sf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{6 \pm \sqrt{-16}}{2} \rightarrow \begin{cases}\sf{x' = \dfrac{6 + 4i}{2} = 3 + 2i\\\\\sf{x'' = \dfrac{6 - 4i}{2} = 3 - 2i\end{cases}}$

$\boxed{\boxed{\sf G = \left\{3 + 2i\:,3-2i\right\}}}$