Question

100 PONTOS!
Os valores de x que satisfazem a sentença:

Answer 1

Resposta:

$\boxed{\mathtt{D}}$

Explicação passo-a-passo:

$\\ \displaystyle \mathsf{\frac{8}{3 - x} \leq 3 + x} \\\\\\ \mathsf{\frac{8}{3 - x} - 3 - x \leq 0} \\\\\\ \mathsf{\frac{8 \cdot 1 - 3 \cdot (3 - x) - x \cdot (3 - x)}{3 - x} \leq 0} \\\\\\ \mathsf{\frac{8 - 9 + 3x - 3x + x^2}{3 - x} \leq 0} \\\\\\ \mathsf{\frac{x^2 - 1}{3 - x} \leq 0}$

Resolvendo a Inequação Quociente...

NUMERADOR:

$\\ \displaystyle \mathsf{x^2 - 1 \leq 0} \\\\ \mathsf{(x + 1)(x - 1) \leq 0} \\\\ \Rightarrow \boxed{\mathsf{S_{numerador} = \left \{ x \in \mathbb{R} \, / \, - 1 \leq x \leq 1 \right \}}}$

DENOMINADOR:

$\\ \displaystyle \mathsf{3 - x < 0} \\\\ \mathsf{- x < - 3} \\\\ \Rightarrow \boxed{\mathsf{S_{denominador} = \left \{ x \in \mathbb{R} \, / \, x > 3 \right \}}}$

Por fim, estudamos os sinais...

__+___[- 1]___-____[+ 1]__+_______+____

__+________+_________+___(3)__-_____

__+___[- 1]__-_____[+ 1]__+___(3)__-_____

Portanto, $\\ \displaystyle \boxed{\boxed{\mathtt{S = \left \{ x \in \mathbb{R} \, / \, - 1 \leq x \leq 1 \, \cup \, x > 3 \right \}}}}$.