Question

Resolva a inequação nos IR :

${\Huge \bf | 2 - \frac{1}{x} | \leq 5 }$

Answer 1

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$\large\begin{array}{l} \textsf{Resolver a inequa\c{c}\~ao modular:}\\\\ \mathsf{\left|2-\dfrac{\,1\,}{x}\right|\le 5\qquad\quad(com~x\ne 0)}\\\\ \mathsf{-5\le 2-\dfrac{\,1\,}{x}\le 5} \end{array}$


$\large\begin{array}{l} \textsf{Somando }\mathsf{-2}\textsf{ a todos os membros da dupla desigualdade,}\\\\ \mathsf{-5-2\le 2-\dfrac{\,1\,}{x}-2\le 5-2}\\\\ \mathsf{-7\le -\,\dfrac{\,1\,}{x}\le 3} \end{array}$


$\large\begin{array}{l} \textsf{Multiplicando por }\mathsf{-1,}\textsf{ que \'e negativo, o sentido se inverte:}\\\\ \mathsf{-3\le \dfrac{\,1\,}{x}\le 7}\\\\\\ \left\{\! \begin{array}{rclc} \mathsf{-3}&\!\!\le\!\!&\mathsf{\dfrac{1}{x}}&\quad\mathsf{(i)}\\\\ \mathsf{\dfrac{\,1\,}{x}}&\!\!\le\!\!&\mathsf{7}&\quad\mathsf{(ii)} \end{array} \right. \end{array}$


$\large\textsf{Resolvendo separadamente cada inequa\c{c}\~ao}:$


•   $\large\textsf{Inequa\c{c}\~ao (i):}$

$\large\begin{array}{l} \mathsf{-3\le \dfrac{\,1\,}{x}}\\\\ \mathsf{\dfrac{\,1\,}{x}+3\ge 0}\\\\ \mathsf{\dfrac{1+3x}{x}\ge 0}\quad\longleftarrow\quad\textsf{inequa\c{c}\~ao-quociente} \end{array}$


$\large\begin{array}{l} \textsf{Fazendo o quadro de sinais:}\\\\ \begin{array}{cc} \mathsf{1+3x}\quad&\quad\mathsf{\underline{~~---}\underset{-\frac{1}{3}}{\bullet}\underline{++++}\underset{0}{\circ}\underline{+++~~}_\blacktriangleright}\\\\ \mathsf{x}\quad&\quad\mathsf{\underline{~~---}\underset{-\frac{1}{3}}{\bullet}\underline{----}\underset{0}{\circ}\underline{+++~~}_\blacktriangleright}\\\\\\ \mathsf{\dfrac{1+3x}{x}}\quad&\quad\mathsf{\underline{~~+++}\underset{-\frac{1}{3}}{\bullet}\underline{----}\underset{0}{\circ}\underline{+++~~}_\blacktriangleright} \end{array} \end{array}$


$\large\begin{array}{l} \textsf{Como queremos o quociente seja }\mathsf{\ge 0,}\textsf{ o intervalo de interesse \'e}\\\\ \mathsf{x\le -\,\dfrac{\,1\,}{3}~~ou~~x>0}\\\\\\ \mathsf{S_{(i)}=\left]-\infty,\,-\frac{1}{3}\right]\,\cup\,\left]0,\,+\infty\right[.} \end{array}$


•   $\large\textsf{Inequa\c{c}\~ao (ii):}$

$\large\begin{array}{l} \mathsf{\dfrac{\,1\,}{x}\le\mathsf{7}}\\\\ \mathsf{\dfrac{\,1\,}{x}-7\le 0}\\\\ \mathsf{\dfrac{1-7x}{x}\le 0}\quad\longleftarrow\quad\textsf{inequa\c{c}\~ao-quociente} \end{array}$


$\large\begin{array}{l} \textsf{Fazendo o quadro de sinais:}\\\\ \begin{array}{cc} \mathsf{1-7x}\quad&\quad\mathsf{\underline{~~+++}\underset{0}{\circ}\underline{++++}\underset{\frac{1}{7}}{\bullet}\underline{---~~}_\blacktriangleright}\\\\ \mathsf{x}\quad&\quad\mathsf{\underline{~~---}\underset{0}{\circ}\underline{++++}\underset{\frac{1}{7}}{\bullet}\underline{+++~~}_\blacktriangleright}\\\\\\ \mathsf{\dfrac{1-7x}{x}}\quad&\quad\mathsf{\underline{~~---}\underset{0}{\circ}\underline{++++}\underset{\frac{1}{7}}{\bullet}\underline{---~~}_\blacktriangleright} \end{array} \end{array}$


$\large\begin{array}{l} \textsf{Como queremos o quociente seja }\mathsf{\le 0,}\textsf{ o intervalo de interesse \'e}\\\\ \mathsf{x<0~~ou~~x\ge \dfrac{1}{7}}\\\\\\ \mathsf{S_{(ii)}=\left]-\infty,\,0\right[\,\cup\,\left[\frac{1}{7},\,+\infty\right[.} \end{array}$

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$\large\begin{array}{l} \textsf{Agora, faremos a interse\c{c}\~ao entre as solu\c{c}\~oes das inequa\c{c}\~oes}\\\textsf{(i) e (ii):}\\\\\\ \begin{array}{cc} \mathsf{S_{(i)}}\quad&\quad\mathsf{\underline{~~\cdots\cdots\cdots}\underset{-\frac{1}{3}}{\bullet}\underline{\qquad\qquad~}\underset{0}{\circ}\underline{\cdots\cdots}\underset{\frac{1}{7}}{\bullet}\underline{\cdots\cdots\cdots~~}_\blacktriangleright}\\\\ \mathsf{S_{(ii)}}\quad&\quad\mathsf{\underline{~~\cdots\cdots\cdots}\underset{-\frac{1}{3}}{\bullet}\underline{\cdots\cdots\cdots\cdot}\underset{0}{\circ}\underline{\qquad~~}\underset{\frac{1}{7}}{\bullet}\underline{\cdots\cdots\cdots~~}_\blacktriangleright}\\\\\\ \mathsf{S_{(i)}\cap S_{(ii)}}\quad&\quad\mathsf{\underline{~~\cdots\cdots\cdots}\underset{-\frac{1}{3}}{\bullet}\underline{\qquad\qquad~}\underset{0}{\circ}\underline{\qquad~~}\underset{\frac{1}{7}}{\bullet}\underline{\cdots\cdots\cdots~~}_\blacktriangleright} \end{array} \end{array}$


$\large\begin{array}{l} \textsf{A solu\c{c}\~ao para a inequa\c{c}\~ao dada inicialmente \'e}\\\\ \mathsf{S=S_{(i)}\cap S_{(ii)}}\\\\ \mathsf{S=\left\{x\in\mathbb{R}:~x\le -\frac{1}{3}~~ou~~x\ge \frac{1}{7} \right \}} \end{array}$


$\large\begin{array}{l} \textsf{ou usando a nota\c{c}\~ao de intervalos,}\\\\ \mathsf{S=\left]-\infty,\,-\frac{1}{3}\right]\cup\left[\frac{1}{7},\,+\infty \right [.} \end{array}$

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$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$


Tags: inequação modular módulo inequação quociente conjunto interseção solução resolver álgebra