Question

Resolva , em IR , a inequação abaixo :

${\huge \bf \frac{x+1}{|2x-1|} \leq 2 }$

Answer 1

Nós podemos imaginar o problema desta forma:

$\mathsf{\dfrac{f(x)}{g(x)} \leq 2}$

Tomando $\mathsf{h(x)=|2x-1|}$, vamos multiplicar ambos os lados da inequação por h(x):

$\mathsf{\dfrac{x+1}{|2x-1|} \leq 2}\\\\\\\mathsf{x+1 \leq 2\cdot|2x-1|}\\\\\\\mathsf{\dfrac{x+1}{2} \leq |2x-1|}$

Pela propriedade do módulo para um número real k, k > 0:

$\mathsf{|x| \geq \ k~\Longrightarrow~x \leq -k~~~~ou~~~~x \geq k}$

$\mathsf{\dfrac{x+1}{2} \leq |2x-1|~\Longrightarrow~2x-1 \leq \dfrac{-x-1}{~~2}~~ou~~{2x-1 \geq \dfrac{x+1}{2}}}$

Então:

$\mathsf{2x-1 \leq \dfrac{-x-1}{~2}}\\\\\mathsf{4x-2 \leq -x-1}\\\\\mathsf{5x \leq 1}\\\\\mathsf{x \leq \dfrac{1}{5}~~~~S_1}\\\\-----------------------------\\\\\mathsf{2x-1 \geq \dfrac{x+1}{2}}\\\\\mathsf{4x-2 \geq x+1}\\\\\mathsf{3x \geq 3}\\\\\mathsf{x \geq 1~~~~S_2}\\\\-----------------------------\\\\\mathsf{S_1\cup S_2=x \leq \dfrac{1}{5}~~ou~~x \geq 1}\\\\-----------------------------$

$\mathsf{S_1\cup S_2=\hspace{20}_{_\blacktriangleleft}\underline{\hspace{........................}}\underset{\frac{1}{5}}\bullet\underline{\hspace{30}}\underset{1}\bullet\underline{\hspace{........................}}_{_\blacktriangleright}}$