Question

Resolva as inequações em R

(2x-1) (-3x-2)
------------------ >= 0
(-4x+5)

Answer 1

$\large\boxed{\begin{array}{l}\sf\dfrac{(2x-1)\cdot(-3x-2)}{-4x+5}\geqslant0\\\underline{\rm fac_{\!\!,}a}\\\sf f(x)=2x-1,g(x)=-3x-2~e~h(x)=-4x+5\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Ra\acute izes\,de\,f(x):}\\\sf 2x-1=0\\\sf 2x=1\implies x=\dfrac{1}{2}\\\underline{\rm Estudo\,do\,sinal\!:}\\\sf f(x)>0\implies x>\dfrac{1}{2}\\\\\sf f(x)<0\implies x<\dfrac{1}{2}\\\underline{\rm Ra\acute izes\,de\,g(x)\!:}\\\sf-3x-2=0\\\sf 3x=-2\implies x=-\dfrac{2}{3}\\\underline{\rm Estudo\,do\,sinal\!:}\\\\\sf g(x)>0\implies x<-\dfrac{2}{3}\\\\\sf g(x)<0\implies x>-\dfrac{2}{3}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Ra\acute izes\,de\,h(x)\!:}\\\sf -4x+5=0\\\sf 4x=5\implies x=\dfrac{5}{4}\\\underline{\rm Estudo\,do\,sinal\!:}\\\sf h(x)>0\implies x<\dfrac{5}{4}\\\\\sf h(x)<0\implies x>\dfrac{5}{4}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Observe\,o\,quadro\,sinal\,que\,anexei.}\\\sf Assinalando\,a\,resposta\,temos:\\\sf S=\bigg\{ x\in\mathbb{R}/-\dfrac{2}{3}\leqslant x\leqslant\dfrac{1}{2}~~ou~x>\dfrac{5}{4}\bigg\}\end{array}}$