Question

Como resolver essa Inequação do Quociente?
1. . . . . . 2. . . .3
___ + ___ - ___ < 0
x -1.... x- 2.... x - 3

Answer 1

$\large\begin{array}{l} \textsf{Resolver a inequa\c{c}\~ao-quociente}\\\\ \mathsf{\dfrac{1}{x-1}+\dfrac{2}{x-2}-\dfrac{3}{x-3}<0} \end{array}$


$\large\textsf{Primeiro, reduza as fra\c{c}\~oes do lado esquerdo ao mesmo}\\\textsf{denominador:}$

$\mathsf{\dfrac{1(x-2)(x-3)}{(x-1)(x-2)(x-3)}+\dfrac{2(x-1)(x-3)}{(x-1)(x-2)(x-3)}-\dfrac{3(x-1)(x-2)}{(x-1)(x-2)(x-3)}<0}\\\\\\ \mathsf{\dfrac{1(x-2)(x-3)+2(x-1)(x-3)-3(x-1)(x-2)}{(x-1)(x-2)(x-3)}<0}\\\\\\ \mathsf{\dfrac{1(x^2-3x-2x+6)+2(x^2-3x-x+3)-3(x^2-2x-x+2)}{(x-1)(x-2)(x-3)}<0}$

$\large\begin{array}{l} \mathsf{\dfrac{1(x^2-5x+6)+2(x^2-4x+3)-3(x^2-3x+2)}{(x-1)(x-2)(x-3)}<0}\\\\ \mathsf{\dfrac{x^2-5x+6+2x^2-8x+6-3x^2+9x-6}{(x-1)(x-2)(x-3)}<0}\\\\ \mathsf{\dfrac{x^2+2x^2-3x^2-5x-8x+9x+6+6-6}{(x-1)(x-2)(x-3)}<0}\\\\ \mathsf{\dfrac{-4x+6}{(x-1)(x-2)(x-3)}<0}\\\\ \mathsf{\dfrac{-2(2x-3)}{(x-1)(x-2)(x-3)}<0} \end{array}$


$\large\begin{array}{l} \textsf{Dividindo os dois lados por }\mathsf{(-2),}\textsf{ que \'e negativo, o sentido}\\\textsf{da desigualdade se inverte:}\\\\ \textsf{(o sinal \textless~torna-se \textgreater)}\\\\ \mathsf{\dfrac{2x-3}{(x-1)(x-2)(x-3)}>0\qquad(ii)} \end{array}$


$\large\begin{array}{l} \textsf{Vamos lembrar que o denominador da fra\c{c}\~ao n\~ao pode se}\\\textsf{anular.}\\\\\textsf{Deve-se ter necessariamente,}\\\\ \mathsf{x\ne 1~~e~~x\ne 2~~e~~x\ne 3} \end{array}$


$\large\begin{array}{l} \textsf{Agora, analisamos os sinais de cada fator envolvido:}\\\\ \begin{array}{cc} \mathsf{(2x-3)}&\mathsf{\underline{~--}\underset{1}{\circ}\underline{--}\underset{\frac{3}{2}}{\bullet}\underline{++}\underset{2}{\circ}\underline{+++}\underset{3}{\circ}\underline{++~}}\\ \mathsf{(x-1)}&\mathsf{\underline{~--}\underset{1}{\circ}\underline{++}\underset{\frac{3}{2}}{\bullet}\underline{++}\underset{2}{\circ}\underline{+++}\underset{3}{\circ}\underline{++~}}\\ \mathsf{(x-2)}&\mathsf{\underline{~--}\underset{1}{\circ}\underline{--}\underset{\frac{3}{2}}{\bullet}\underline{--}\underset{2}{\circ}\underline{+++}\underset{3}{\circ}\underline{++~}}\\\mathsf{(x-3)}&\mathsf{\underline{~--}\underset{1}{\circ}\underline{--}\underset{\frac{3}{2}}{\bullet}\underline{--}\underset{2}{\circ}\underline{---}\underset{3}{\circ}\underline{++~}}\\\\ \mathsf{\dfrac{2x-3}{(x-1)(x-2)(x-3)}}&\mathsf{\underline{~++}\underset{1}{\circ}\underline{--}\underset{\frac{3}{2}}{\bullet}\underline{++}\underset{2}{\circ}\underline{---}\underset{3}{\circ}\underline{++~}}\end{array} \end{array}$


$\large\begin{array}{l} \textsf{Queremos que o quociente do lado esquerdo da inequa\c{c}\~ao}\\\textsf{(ii) seja positivo. Ent\~ao, o intervalo de interesse \'e}\\\\ \mathsf{x<1~~ou~~\frac{3}{2}<x<2~~ou~~x>3}\\\\\\ \textsf{Conjunto solu\c{c}\~ao: }\mathsf{S=\left\{x\in\mathbb{R}:~x<1~~ou~~\frac{3}{2}<x<2~~ou~~x>3\right\}}\\\\\\ \textsf{ou usando a nota\c{c}\~ao de intervalos,}\\\\ \mathsf{S=\left]-\infty,\,1\right[\,\cup\,\left]\frac{3}{2},\,2\right[\,\cup\,\left]3,\,+\infty\right[\,.} \end{array}$


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


Tags: inequação quociente produto algébrica desigualdade sinal álgebra