Colégio Naval 1979 ⇒ Questão 15 - CN - 1979 Tópico resolvido

[petrasMOD]

15) Simplificando [tex3]\frac{(2x^2-4x+8)(x^2-4)}{\sqrt{2}x^3+\sqrt{128}}[/tex3] vamos encontrar:
a) [tex3]\sqrt{2}[/tex3] (x +2) b) [tex3]\sqrt{2}[/tex3] (x -2) c) [tex3]\sqrt{2}[/tex3] (x² -4) d) [tex3]\sqrt{2}[/tex3] e) 2[tex3]\sqrt{2}[/tex3]

[Kin07]

Resolução:
[tex3]\large \displaystyle \sf \dfrac{(2x^2-4x+8)\cdot (x^2-4)}{\sqrt{2}x^3+\sqrt{128}} [/tex3]

Fatorando as expressões, temos:
[tex3]\large \displaystyle \sf \dfrac{(\,2x^2-4x+8\,)\cdot [\, (\, x+2 \,) \cdot (\, x-2\,)\,]}{\sqrt{2}x^3+\sqrt{64 \cdot 2}} [/tex3]

[tex3]\large \displaystyle \sf \dfrac{ 2\cdot (\,x^2-2x+4 \,)\cdot [\, (\, x+2 \,) \cdot (\, x-2\,)\,]}{\sqrt{2}x^3+\sqrt{64} \cdot \sqrt{2}} [/tex3]

[tex3]\large \displaystyle \sf \dfrac{ 2\cdot (\,x^2-2x+4 \,)\cdot [\, (\, x+2 \,) \cdot (\, x-2\,)\,]}{\sqrt{2}x^3+ 8 \cdot \sqrt{2}} [/tex3]

[tex3]\large \displaystyle \sf \dfrac{ 2\cdot (\,x^2-2x+4 \,)\cdot [\, (\, x+2 \,) \cdot (\, x-2\,)\,]}{\sqrt{2} \cdot (\,x^3+ 8 \,)} [/tex3]

[tex3]\large \displaystyle \sf \dfrac{ 2\cdot (\,x^2-2x+4 \,)\cdot [\, (\, x+2 \,) \cdot (\, x-2\,)\,]}{\sqrt{2} \cdot [\,(x + 2) \cdot (x^2 - 2x + 4)\,] } [/tex3]

[tex3]\large \displaystyle \sf \dfrac{ 2\cdot \cancel{ (\,x^2-2x+4 \,)}\cdot [\, \cancel{(\, x+2 \,)} \cdot (\, x-2\,)\,]}{\sqrt{2} \cdot [\, \cancel{(x + 2)} \cdot \cancel{(x^2 - 2x + 4)}\,] } [/tex3]

[tex3] \large \displaystyle \sf \dfrac{ 2 \cdot (\, x-2\,)}{ \sqrt{2}} \times \dfrac{ \sqrt{2} }{ \sqrt{2}} [/tex3]

[tex3] \large \displaystyle \sf \dfrac{ 2 \cdot \sqrt{2} \cdot (\, x-2\,)}{ \sqrt{4}} [/tex3]

[tex3] \large \displaystyle \sf \dfrac{ 2 \cdot \sqrt{2} \cdot (\, x-2\,)}{ 2} [/tex3]

[tex3] \large \displaystyle \sf \dfrac{ \cancel{2} \cdot \sqrt{2} \cdot (\, x-2\,)}{ \cancel{2}} [/tex3]

[tex3] \large \displaystyle \colorbox{#FFBF00}{ $ \sf \sqrt{2}\cdot (\, x-2\,) $ } [/tex3]

Alternativa correta é a letra B.