Question

Resolva,em R, as inequações-quociente:
A) X-3/2x-1 > = 4
B)X+1/2x-1 < = 0

P.S. : Os simbolos > = e < = significa (Maior igual ou menor igual)
Pfv me ajudem!!!

Answer 1

Resolver a inequação do tipo quociente

$\mathtt{\dfrac{x-3}{2x-1}\le 4}\quad\quad\texttt{com }\mathtt{x\ne \dfrac{1}{2}}$


Primeiro passamos todos os termos para o mesmo membro da desigualdade

$\mathtt{\dfrac{x-3}{2x-1}-4\le 0}$


E depois reduzimos os termos ao mesmo denominador

$\mathtt{\dfrac{x-3}{2x-1}-\dfrac{4\cdot (2x-1)}{2x-1}\le 0}\\\\\\ \mathtt{\dfrac{x-3}{2x-1}-\dfrac{8x-4}{2x-1}\le 0}\\\\\\ \mathtt{\dfrac{x-3-(8x-4)}{2x-1}\le 0}\\\\\\ \mathtt{\dfrac{x-3-8x+4}{2x-1}\le 0}\\\\\\ \mathtt{\dfrac{-7x+1}{2x-1}\le 0}$


Podemos dizer que temos

$\mathtt{\dfrac{f(x)}{g(x)}\le 0}$


Sendo

$\mathtt{f(x)=-7x+1~~e~~g(x)=2x-1.}$

__________


Agora vamos estudar os sinais de $\mathtt{f}$ e $\mathtt{g}$

•   $\mathtt{f(x)=-7x+1}$


$\mathtt{f(x)\geq 0}\\\\ \mathtt{-7x+1\geq 0}\\\\ \mathtt{-7x\geq -1}\\\\ \mathtt{x \leq \dfrac{1}{7}}$


Portanto

$\begin{cases} \mathtt{f(x)\geq 0}&\texttt{se }\mathtt{x\leq 1/7}\\ \mathtt{f(x) \ \textless \ 0}&\texttt{se }\mathtt{x\ \textgreater \ 1/7 } \end{cases}$

$\begin{array}{cc} \mathtt{f(x)}~&~\mathtt{\underline{++}\underset{1/7}{\circ}\underline{----}} \end{array}$


•   $\mathtt{g(x)=2x-1}$


$\mathtt{g(x)\geq 0}\\\\ \mathtt{2x-1\geq 0}\\\\ \mathtt{2x\geq 1}\\\\ \mathtt{x \ \textgreater \ \dfrac{1}{2}}$


Portanto

$\begin{cases} \mathtt{g(x)\geq 0}&\texttt{se }\mathtt{x \geq 1/2}\\ \mathtt{g(x) \ \textless \ 0}&\texttt{se }\mathtt{x\ \textless \ 1/2 } \end{cases}$

$\begin{array}{cc} \mathtt{g(x)}~&~\mathtt{\underline{----}\underset{1/2}\bullet \underline{++}} \end{array}$

__________


$\begin{array}{cc} \mathtt{f(x)}~&~\mathtt{\underline{++++}\underset{1/7}\bullet \underline{--------}} \end{array}\\\\ \begin{array}{cc} \mathtt{g(x)}~&~\mathtt{\underline{----------}\underset{1/2}{\circ}\underline{++++}} \end{array}\\\\\\ \begin{array}{cc} \mathtt{\dfrac{f(x)}{g(x)}}~&~\mathtt{\underline{----}\underset{1/7}\bullet\underline{++++}\underset{1/2}{\circ} \underline{----}} \end{array}$


Obs: lembrando que $\mathtt{x \neq 2}$ porque caso contrário o denominador será zero


Conjunto solução: $\mathsf{S=[\frac{1}{7},\frac{1}{2}[}$


===================


A segunda já é bem mais simples

$\mathtt{\dfrac{x+1}{2x-1}\le 4}\quad\quad\texttt{com }\mathtt{x\ne \dfrac{1}{2}}$


Para o numerador temos

$\mathtt{x+1 \leq 0}\\\\ \mathtt{x\leq -1}$


E para o denominador

$\mathtt{2x-1 \leq 0}\\\\ \mathtt{2x \leq 1}\\\\ \mathtt{x \leq \dfrac{1}{2} }$


Fazendo a análise dos sinais

$\begin{array}{cc} \mathtt{N}~&~\mathtt{\underline{----}\underset{-1}\bullet \underline{++++++++}} \end{array}\\\\ \begin{array}{cc} \mathtt{D}~&~\mathtt{\underline{--------}\underset{1/2}{\circ}\underline{++++}} \end{array}\\\\\\ \begin{array}{cc} \mathtt{Q}~&~\mathtt{\underline{++++}\underset{-1}\bullet \underline{---}\underset{1/2}{\circ} \underline{++++}} \end{array}$


Obs: novamente $\mathtt{x \neq \dfrac{1}{2}}$ senão o denominador ficará 0


Conjunto solução: $\mathtt{S=[-1,\dfrac{1}{2}[}$


Bons estudos no Brainly! =)