Question

Resolva, em ℝ, as inequações. (cálculos)

A) (3x − 9)(3 − x) > 0

B)
3x − 6/

5 − x > 0

C)
x + 1/

2 − 3x ≤ 0​


por favor me ajude

Answer 1

$\large\boxed{\begin{array}{l}\tt a)~\sf(3x-9)\cdot(3-x)>0\\\sf interprete~cada~parcela~da~inequac_{\!\!,}\tilde ao-produto\\\sf como~func_{\!\!,}\tilde oes,em~seguida~fac_{\!\!,}a~o~estudo~do~sinal\\\sf por~fim, elabore~um~quadro~de~sinais~e~assinale\\\sf o~intervalo~positivo.\end{array}}$

$\boxed{\begin{array}{l}\sf f(x)=3x-9\\\underline{\rm ra\acute izes~de~f(x):}\\\sf 3x-9=0\\\sf 3x=9\\\sf x=\dfrac{9}{3}\\\\\sf x=3\\\sf f(x)>0~para~x>3~e~f(x)<0~para~x<3\\\sf g(x)=3-x\\\underline{\rm ra\acute izes~de~g(x):}\\\sf 3-x=0\\\sf x=3\\\sf g(x)>0~para~x<3~e~g(x)<0~para~x>3\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm Observe~a~figura~que~eu~anexei.}\\\sf Perceba~que~n\tilde ao~temos~intervalos~positivos\\\sf para~o~produto~destas~func_{\!\!,}\tilde oes\\\sf portanto\\\sf S=\bigg\{\bigg\}\end{array}}$

$\boxed{\begin{array}{l}\tt b)~\sf\dfrac{3x-6}{5-x}>0\\\\\sf f(x)=3x-6\\\underline{\rm Ra\acute izes~de~f(x):}\\\sf 3x-6=0\\\sf 3x=6\\\sf x=\dfrac{6}{3}\\\\\sf x=2\\\sf f(x)>0~para~x>2~f(x)<0~para~x<2\\\sf g(x)=5-x\\\underline{\rm Ra\acute izes~de~g(x):}\\\sf 5-x=0\\\sf x=5\\\sf g(x)>0~para~x<5~e~g(x)<0~para~x>5\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Observe~anexo~3.}\\\sf Assinalando~o~trecho~positivo~temos\\\sf S=\{x\in\mathbb{R}/2<x<5\}\end{array}}$

$\large\boxed{\begin{array}{l}\tt c)~\sf\dfrac{x+1}{2-3x}\leqslant0\\\\\sf f(x)=x+1\\\underline{\rm Ra\acute izes~de~f(x):}\\\sf x+1=0\\\sf x=-1\\\sf f(x)>0~para~x>-1~e~f(x)<0~para~x<-1\\\sf g(x)=2-3x\\\underline{\rm Ra\acute izes~de~g(x):}\\\sf 2-3x=0\\\sf 3x=2\\\sf x=\dfrac{2}{3}\\\\\sf g(x)>0~para~x<\dfrac{2}{3}~e~g(x)<0~para~x>\dfrac{2}{3}\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Observe~anexo~2}\\\sf Assinalando~o~intervalo~negativo~temos\\\sf S=\bigg\{x\leqslant-1~ou~x>\dfrac{2}{3}\bigg\}\end{array}}$