Question

Resolva as seguintes inequações quociente:​

Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm Produto\,de\,Stevin}\\\sf (x+a)\cdot (x+b)=x^2+(a+b)x+a\cdot b\end{array}}$

$\Large\boxed{\begin{array}{l}\tt a)~\sf\dfrac{x^2-7x+10}{x^2-5x+4}>0\\\\\sf f(x)=x^2-7x+10\\\underline{\rm ra\acute izes\,de\,f(x):}\\\sf x^2-7x+10=0\\\sf x^2+(-2-5)+(-2)\cdot( -5)=0\\\sf (x-2)\cdot(x-5)=0\\\sf x-2=0\implies x=2\\\sf x-5=0\implies x=5\\\sf f(x)>0\iff x<2~ou~x>5\\\sf f(x)<0\iff 2<x<5\end{array}}$

$\Large\boxed{\begin{array}{l}\sf g(x)=x^2-5x+4\\\underline{\rm ra\acute izes\,de\,g(x)}\\\sf x^2-5x+4=0\\\sf x^2+(-1-4)x+(-1)\cdot(-4)=0\\\sf(x-1)\cdot(x-4)=0\\\sf x-1=0\implies x=1\\\sf x-4=0\implies x=4\\\sf g(x)>0\iff x<1~ou~x<4\\\sf g(x)<0\iff 1<x<4\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Observando\,o\,quadro\,sinal}\\\underline{\rm no\,anexo\,1\,temos:}\\\sf S=\{x\in\mathbb{R}/x<1~ou~2<x<4~ou~x>5\}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt b)~\sf\dfrac{x^2+2}{x^2-3x}\leqslant0\\\\\sf f(x)=x^2+2\\\underline{\rm ra\acute izes\,de\,f(x)}\\\sf x^2+2=0\\\sf x^2=-2\implies \not\exists\,x\in\mathbb{R}\\\sf f(x)>0,\forall x\in\mathbb{R}\\\sf g(x)=x^2-3x\\\underline{\rm ra\acute izes\,de\,g( x)}\\\sf x^2-3x=0\\\sf x\cdot(x-3)=0\\\sf x=0\\\sf x-3=0\\\sf x=3\\\sf g(x)>0\iff x<0~ou~x>3\\\sf g(x)<0\iff 0<x<3\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Observe\,o\,quadrado\,sinal}\\\underline{\rm no\,anexo\,2\,temos:}\\\sf S=\{x\in\mathbb{R}/0<x<3\}\end{array}}$