Question

Resolva as inequações​

Answer 1

Resposta:

a)

$\sf 2^{x +1} < 2$

$\sf 2^{x +1} < 2^1$

$\sf x + 1 < 1$

$\sf x < 1 - 1$

$\sf x < 0$

$\boldsymbol{ \sf \displaystyle S=\{x\in\mathbb{R} \mid x < 0 \} }$

b)

$\sf \left ( \dfrac{1}{2} \right )^{x + 1} \ge 8^{x + 2}$

$\sf \left ( 2^{(-1)})^{x + 1} \ge 2^3^{(x + 2)}$     ←  Cancelar a base 2.

$\sf -\: x - 1 \ge 3x + 6$

$\sf -x -3x \ge 6 +1$

$\sf - 4x \ge 7$    ←  Multiplicar por ( -1), lembrando que inverte o sinal.

$\sf 4x \le - 7$

$\sf x \le - \dfrac{7}{4}$

$\boldsymbol{ \sf \displaystyle S= \bigg \{x\in\mathbb{R} \mid x \le - \dfrac{7}{4} \bigg\} }$

Explicação passo-a-passo: