Question

(X^2 - 5x + 6)^3 - ( x^2 - 5x + 6) = 0

Answer 1

Olá! Boa noite.

$\\ \mathsf{(x^2 - 5x + 6)^3 - (x^2 - 5x + 6) = 0} \\\\ \mathsf{(x^2 - 5x + 6) \cdot \left [ (x^2 - 5x + 6)^2 - 1 \right ] = 0} \\\\ \mathsf{(x^2 - 5x + 6) \cdot \left \{ \left [ (x^2 - 5x + 6) + 1 \right ] \cdot \left [ (x^2 - 5x + 6) - 1 \right ] \right \} =} \\\\ \mathsf{(x^2 - 5x + 6) \cdot (x^2 - 5x + 7) \cdot (x^2 - 5x + 5) = 0} \\\\ \mathsf{\underbrace{\mathsf{(x^2 - 5x + 5)}}_{(I)}\underbrace{\mathsf{(x^2 - 5x + 6)}}_{(II)}\underbrace{\mathsf{(x^2 - 5x + 7)}}_{(III)} = 0}$

 Resolvendo cada fator por Bháskara:

Fator I:

$\\ \mathsf{\Delta = 25 - 20 = 5} \\\\ \mathsf{x = \frac{5 \pm \sqrt{5}}{2}} \\\\ \boxed{\mathsf{x_1 = \frac{5 + \sqrt{5}}{2}}} \ \text{e} \ \boxed{\mathsf{x_2 = \frac{5 - \sqrt{5}}{2}}}$

Fator II:

$\\ \mathsf{\Delta = 25 - 24 = 1} \\\\ \mathsf{x = \frac{5 \pm \sqrt{1}}{2}} \\\\ \boxed{\mathsf{x_3 = 3}} \ \text{e} \ \boxed{\mathsf{x_4 = 2}}$

Fator III:

$\\ \mathsf{\Delta = 25 - 28 = - 3 = 3i^2} \\\\ \mathsf{x = \frac{5 \pm i\sqrt{3}}{2}} \\\\ \boxed{\mathsf{x_5 = \frac{5 + i\sqrt{3}}{2}}} \ \text{e} \ \boxed{\mathsf{x_6 = \frac{5 - i\sqrt{3}}{2}}}$

 
 Logo, $\boxed{\boxed{\mathsf{S = \left \{ 2, 3, \frac{5 \pm \sqrt{5}}{2}, \frac{5 \pm i\sqrt{3}}{2} \right \}}}}$

Answer 2

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Explicação passo a passo:

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