Question

resolva o sistema

$\left \{ {x²-3x+2\ \textgreater \ 0} \atop {-x²+x\ \textgreater \ 0}} \right.$

Answer 1

Caso tenha problemas para visualizar a resposta experimente abrir pelo navegador https://brainly.com.br/tarefa/34488049

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

$\begin{cases}\sf x^2-3x+2>0\\\sf -x^2+x>0\end{cases}\\\sf f(x)=x^2-3x+2\\\sf x^2-3x+2=0\\\sf \Delta=b^2-4ac\\\sf \Delta=(-3)^2-4\cdot1\cdot2\\\sf\Delta=9-8\\\sf\Delta=1\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf x=\dfrac{-(-3)\pm\sqrt{1}}{2\cdot1}\\\sf x=\dfrac{3\pm 1}{2}\begin{cases}\sf x_1=\dfrac{3+1}{2}=\dfrac{4}{2}=2\\\sf x_2=\dfrac{3-1}{2}=\dfrac{2}{2}=1\end{cases}\\\sf f(x)>0\implies x<1~~ou~~x>2\\\sf f(x)<0\implies 1<x<2\\\sf s_1=\{x\in\mathbb{R}/ x<1~~ou~~x>2\}$

$\sf g(x)=-x^2+x\\\sf -x^2+x=0\\\sf x\cdot(-x+1)=0\\\sf x=0\\\sf -x+1=0\\\sf x=1\\\sf g(x)>0\implies 0<x<1\\\sf g(x)<0\implies x<0~~ou~~x>1\\\sf s_2=\{x\in\mathbb{R}/0<x<1\}$

$\sf montando~a~reta~real~e~assinalando~a~intersecc_{\!\!,}\tilde{a}o~temos:$

$\sf S=S_1\cap S_2\\\huge\boxed{\boxed{\boxed{\boxed{\sf S=\{x\in\mathbb{R}/0<x<1\}}}}}$