Question

Resolva a inequacao: (2/x) < x-1

Answer 1

Explicação passo-a-passo:

$\sf \dfrac{2}{x} < x-1$

$\sf \dfrac{2}{x}-x+1 < 0$

$\sf \dfrac{2-x^2+x}{x} < 0$

$\sf \dfrac{-x^2+x+2}{x} < 0$

• $\sf -x^2+x+2=0$

$\sf \Delta=1^2-4\cdot(-1)\cdot2$

$\sf \Delta=1+8$

$\sf \Delta=9$

$\sf x=\dfrac{-1\pm\sqrt{9}}{2\cdot(-1)}=\dfrac{-1\pm3}{-2}$

$\sf x'=\dfrac{-1+3}{-2}~\Rightarrow~x'=\dfrac{2}{-2}~\Rightarrow~x'=-1$

$\sf x"=\dfrac{-1-3}{-2}~\Rightarrow~x"=\dfrac{-4}{-2}~\Rightarrow~x"=2$

Temos que:

$\sf ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-1~~~~~~~0~~~~~~~~2~~$

$\sf -x^2+x+2~~~~~~~|~~-~~|~~+~~|~~+~~|~~-~~|$

$\sf ~~~~~~~~~~x~~~~~~~~~~~~~~~|~~-~~|~~-~~|~~+~~|~~+~~|$

$\sf \dfrac{-x^2+x+2}{x}~~~~~~|~~+~~|~~-~~|~~+~~|~~-~~|$

Logo, $\sf S=\{x\in\mathbb{R}~|-1 < x < 0~ou~x>2\}$

Answer 2

Oie, Td Bom?

2/x < x - 1

... Encontre os valores restritos para x que façam com o que o denominador de 2/x seja igual a 0.

• x = 0

2/x < x - 1 , x ≠ 0

2/x - x + 1 < 0

2/x - x/1 + 1/1 < 0

2/x - [x . x]/[x . 1] + [x . 1]/[x . 1] < 0

2/x - x²/x + x/x < 0

[2 - x² + x]/x < 0

... Divida em casos possíveis.

{2 - x² + x < 0 ⇒ x∈⟨- ∞ , - 1⟩∪⟨2 , + ∞⟩

{x > 0

{2 - x² + x > 0 ⇒ x∈⟨- 1 , 2⟩

{x < 0

... Encontre a interseção.

x∈⟨2 , + ∞⟩

x∈⟨- 1 , 0⟩

... Encontre a união.

x∈⟨- 1 , 0⟩∪⟨2 , + ∞⟩ , x ≠ 0

x∈⟨- 1 , 0⟩∪⟨2 , + ∞⟩

Att. Makaveli1996