Question

Resolva, em IR, as inequações.

Answer 1

$\large\boxed{\begin{array}{l}\underline{\sf Desigualdades\,envolvendo\,m\acute odulo}\\\rm |x|<a\iff -a<x<a\\\rm |x|>a\iff x<-a~~ou~~x>a\end{array}}$

$\large\boxed{\begin{array}{l}\sf a)~\rm|5x+7|\geqslant13\\\rm 5x+13\leqslant-13\\\rm 5x\leqslant-13-13\\\rm 5x\leqslant-26\\\rm x\leqslant-\dfrac{26}{5}\\\rm 5x+7\geqslant13\\\rm 5x\geqslant13-7\\\rm 5x\geqslant6\\\rm x\geqslant\dfrac{6}{5}\\\rm S=\bigg\{x\in\mathbb{R}\bigg/x\leqslant-\dfrac{26}{5}~ou~x\geqslant\dfrac{6}{5}\bigg\}\end{array}}$

$\large\boxed{\begin{array}{l}\sf b)~\rm|3x-4|<8\\\rm -8<3x-4<8\\\rm -8+4<3x<8+4\\\rm -4<3x<12\\\\\rm -\dfrac{4}{3}<x<\dfrac{12}{3}\\\\\rm -\dfrac{4}{3}<x<4\end{array}}$

$\large\boxed{\begin{array}{l}\sf\,\rm\bigg|\dfrac{3x}{4}+\dfrac{1}{2}\bigg|\leqslant\dfrac{1}{5}\\\\\rm -\dfrac{1}{5}\leqslant\dfrac{3x}{4}+\dfrac{1}{2}\leqslant\dfrac{1}{5}\bullet(20)\\\\\rm -4\leqslant15x+10\leqslant4\\\rm -4-10\leqslant15x\leqslant4-10\\\rm -14\leqslant15x\leqslant-6\\\\\rm-\dfrac{14}{15}\leqslant x\leqslant-\dfrac{6\div3}{15\div3}\\\\\rm -\dfrac{14}{15}\leqslant x\leqslant-\dfrac{2}{5}\end{array}}$

$\huge\boxed{\boxed{\boxed{\boxed{\rm S=\bigg\{x\in\mathbb{R}\bigg/-\dfrac{14}{15}\leqslant x\leqslant-\dfrac{2}{5}\bigg\}}}}}$