Question

Resolva em IR a inequação:

$\frac{x+2}{3}- \frac{ x^{2}-4 }{2} \ \textgreater \ -1- \frac{5x-2}{6}$

FAÇA COM A RESOLUÇÃO!!

Answer 1

$\mathsf{Ola\´~Fabio,}$

$\mathsf{\frac{x+2}{3}-\frac{x^2-4}{2}\ \textgreater \ -1-\frac{5x-2}{6}}\\\\\mathsf{mmc(3,2,6})=6}\\\\\mathsf{\frac{2x+4}{\not{6}}-\frac{3x^2-12}{\not{6}}\ \textgreater \ \frac{-6}{\not{6}}-\frac{5x-2}{\not{6}}}\\\\\mathsf{2x+4-(3x^2-12)\ \textgreater \ -6-(5x-2)}\\\mathsf{2x+4-3x^2+12\ \textgreater \ -6-5x+2}\\\mathsf{-3x^2+2x+16\ \textgreater \ -5x-4}\\\mathsf{-3x^2+7x+20\ \textgreater \ 0~*(-1)}\\\mathsf{3x^2-7x-20\ \textless \ 0}$

$\Delta=b^2 - 4.a.c\\\Delta=7^2-4.3.(-20)\\\Delta=49+240\\\Delta=289$


$x=\frac{-b\pm\sqrt{delta}}{2a}\\\\x=\frac{-(-7)\pm\sqrt{289}}{2.3}\\\\x=\frac{7+17}{6}\\\\\boxed{x=4}\\\\x'=\frac{7-17}{6}\\\\x'=\frac{-10}{6}\\\\\boxed{x'=-\frac{5}{3}}$


$-\frac{5}{3}~\bigcap~4~ \textgreater \ 0\\\\\\\mathsf{\Large\boxed{S=\{x\in~\mathbb{R}~|~-\frac{5}{3}\ \textless \ x\ \textless \ 4\}}}\\\\\\\\\mathsf{Du\´vidas?~comente}$

Answer 2

$\large{\begin{array}{l} \mathsf{\dfrac{x+2}{3}-\dfrac{x^2-4}{2}>-1-\dfrac{5x-2}{6}} \end{array}}$


$\large{\begin{array}{l} \textsf{Reduza os termos de cada membro ao mesmo denominador:}\\\\ \mathsf{\dfrac{2(x+2)}{6}-\dfrac{3(x^2-4)}{6}>-1-\dfrac{5x-2}{6}}\\\\ \mathsf{\dfrac{2(x+2)}{6}-\dfrac{3(x^2-4)}{6}>\dfrac{-6}{6}-\dfrac{5x-2}{6}}\\\\ \mathsf{\dfrac{2(x+2)-3(x^2-4)}{6}>\dfrac{-6-(5x-2)}{6}} \end{array}}$


$\large{\begin{array}{l} \textsf{Multiplicando os dois lados por 6, que \'e positivo, o}\\\textsf{sentido da desigualdade \'e preservado}\\\\ (\textsf{o sinal}>\textsf{permanece}>)\\\\ \mathsf{\dfrac{2(x+2)-3(x^2-4)}{\diagup\!\!\!\! 6}\cdot \diagup\!\!\!\! 6>\dfrac{-6-(5x-2)}{\diagup\!\!\!\! 6}\cdot \diagup\!\!\!\! 6}\\\\ \mathsf{2(x+2)-3(x^2-4)>-6-(5x-2)}\\\\ \mathsf{2x+4-3x^2+12>-6-5x+2}\\\\ \end{array}}$

$\large{\begin{array}{l} \mathsf{2x+4-3x^2+12+6+5x-2>0}\\\\ \mathsf{-3x^2+2x+5x+4+12+6-2>0}\\\\ \mathsf{-3x^2+7x+20>0\qquad\rightarrow\cdot \,(-1)}\\\\ \mathsf{3x^2-7x-20<0\qquad(i)} \end{array}}$


$\large{\begin{array}{l} \textsf{O que podemos fazer \'e achar as ra\'izes do lado esquerdo}\\\textsf{fator\'a-lo e expressar (i) como uma inequa\c{c}\~ao-produto.}\\\\ \textsf{Poder\'iamos achar as ra\'izes resolvedo a equa\c{c}\~ao pela}\\\textsf{f\'ormula de B\'ascara, mas aqui, vou usar fatora\c{c}\~ao por}\\\textsf{agrupamento:} \end{array}}$


$\large{\begin{array}{l} \textsf{Em (i), Reescreva convenientemente }\mathsf{-7x}\textsf{ como }\mathsf{-12x+5x:}\\\\ \mathsf{3x^2-12x+5x-20<0}\\\\ \mathsf{3x(x-4)+5(x-4)<0}\\\\ \mathsf{(x-4)(3x+5)<0}\\\\ \mathsf{3(x-4)(x+\frac{5}{3})<0}\\\\ \mathsf{(x-4)(x+\frac{5}{3})<0\qquad(ii)} \end{array}}$


$\large{\begin{array}{l}\\\\ \textsf{Agora (ii) \'e uma inequa\c{c}\~ao-produto. As ra\'izes do lado}\\\textsf{esquerdo de (ii) s\~ao}\\\\ \mathsf{x_1=-\frac{5}{3}~~e~~x_2=4}\\\\\\ \textsf{Estudando o sinal de cada fator:} \end{array}}$

$\large{\begin{array}{l}\\\\ \begin{array}{cc} \mathsf{(x+\frac{5}{3})}~~&\underline{---}\underset{-\frac{5}{3}}{\bullet}\underline{++++}\underset{4}{\bullet}\underline{++++}\\\\ \mathsf{(x-4)}~~&\underline{---}\underset{-\frac{5}{3}}{\bullet}\underline{----}\underset{4}{\bullet}\underline{++++}\\\\\\ \mathsf{(x-4)(x+\frac{5}{3})}~~&\underline{+++}\underset{-\frac{5}{3}}{\bullet}\underline{----}\underset{4}{\bullet}\underline{++++} \end{array} \end{array}}$


$\large{\begin{array}{l} \textsf{Como queremos que o produto em (ii) seja negativo,}\\\textsf{o intervalo de intersse \'e}\\\\ \mathsf{-\frac{5}{3}<x<4}\\\\\\ \textsf{Conjunto solu\c{c}\~ao: }\mathsf{S=\left\{x\in \mathbb{R}:~-\frac{5}{3}<x<4\right\}}\\\\\\ \textsf{ou usando a nota\c{c}\~ao de intervalos,}\\\\ \mathsf{S=\left]-\frac{5}{3},\,4\right[\,.} \end{array}}$


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


Tags: inequação produto quociente fração desigualdade solução intervalos reais resolver