Determina o conjunto-solução de cada uma das condições
Os números reais: Matematicamente Falando 9 - Parte 1 Pág. 36 Ex. 17
Determina o conjunto-solução de cada uma das condições:
- \({x – 1 \le 7}\)
- \({\left| x \right| > 3}\)
- \({1 < \frac{{x – 3}}{2} \le 7}\)
- Ora,
\[\begin{array}{*{20}{l}}{x – 1 \le 7}& \Leftrightarrow &{x \le 8}\\{}&{}&{}\\{}&{}&{S = \left] { – \infty ,\;8} \right]}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{\left| x \right| > 3}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x < – 3}& \vee &{x > 3}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left] { – \infty ,\; – 3} \right[ \cup \left] {3,\; + \infty } \right[}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{1 < \frac{{x – 3}}{2} \le 7}& \Leftrightarrow &{2 < x – 3 \le 14}\\{}& \Leftrightarrow &{2 + 3 < x – 3 + 3 \le 14 + 3}\\{}& \Leftrightarrow &{5 < x \le 17}\\{}&{}&{}\\{}&{}&{S = \left] {5,\;17} \right]}\end{array}\]
\[\begin{array}{*{20}{l}}{1 < \frac{{x – 3}}{2} \le 7}& \Leftrightarrow &{\begin{array}{*{20}{c}}{1 < \frac{{x – 3}}{2}}& \wedge &{\frac{{x – 3}}{2} \le 7}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{2 < x – 3}& \wedge &{x – 3 \le 14}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x > 5}& \wedge &{x \le 17}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left] {5,\;17} \right]}\end{array}\]
\[\begin{array}{*{20}{l}}{1 < \frac{{x – 3}}{2} \le 7}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{c}}{1 < \frac{{x – 3}}{2}}\\{\frac{{x – 3}}{2} \le 7}\end{array}} \right.}\\{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{c}}{2 < x – 3}\\{x – 3 \le 14}\end{array}} \right.}\\{}& \Leftrightarrow &{\left\{ {\begin{array}{*{20}{c}}{x > 5}\\{x \le 17}\end{array}} \right.}\\{}&{}&{}\\{}&{}&{S = \left] {5,\;17} \right]}\end{array}\]
