Question

como resolvo essa inequação: 1/x+1≥3/x-2

Answer 1

$\Large\boxed{\begin{array}{l}\sf\dfrac{1}{x+1}\geqslant\dfrac{3}{x-2}\\\\\sf\dfrac{1}{x+1}-\dfrac{3}{x-2}\geqslant0\\\\\sf\dfrac{x-2-3\cdot(x+1)}{(x+1)\cdot(x-2)}\geqslant0\\\\\sf\dfrac{x-2-3x-3}{(x+1)\cdot(x-2)}\geqslant0\\\\\sf\dfrac{-2x-5}{(x+1)\cdot(x-2)}\geqslant0\end{array}}$

$\Large\boxed{\begin{array}{l}\sf f(x)=-2x-5\\\underline{\rm ra\acute izes~de~f(x):}\\\sf -2x-5=0\\\sf 2x=-5\\\sf x=-\dfrac{5}{2}\\\sf f(x)>0\implies x<-\dfrac{5}{2}\\\\\sf f(x)<0\implies x>-\dfrac{5}{2}\\\sf g(x)=x+1\\\underline{\rm ra\acute izes~de~g(x):}\\\sf x+1=0\\\sf x=-1\\\sf g(x)>0\implies x>-1\\\sf g(x)<0\implies x<-1\\\sf h(x)=x-2\\\underline{\rm ra\acute izes~de~h(x)}\\\sf x-2=0\\\sf x=2\\\sf h(x)>0\implies x>2\\\sf h(x)<0\implies x<2\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Observe~a~figura~que~eu~anexei}\\\sf Assinalando~o~intervalo~correspondente\\\sf a~parte~positiva~temos:\\\sf S=\bigg\{x\in\mathbb{R}/x\leqslant-\dfrac{5}{2}~ou~-1<x<2\bigg\}\end{array}}$