Question

Resolva, em R, as seguintes inequações:

a) x² + 16 ≥ 0

b) x² + 6x > 0

c) −x² − 4x − 4 < 0


Alguém pode me ajudar pliss?

Answer 1

$\large\boxed{\begin{array}{l}\sf a)\\\rm x^2+16\geqslant0\\\underline{\sf fac_{\!\!,}a}\\\rm f(x)=x^2+16\\\rm f(x)>0\,para\,todo\,x\in\mathbb{R},isto\,\acute e,\\\rm a\,func_{\!\!,}\tilde ao\,nunca\,se\,anula\,portanto\\\rm S=\varnothing\end{array}}$

$\large\boxed{\begin{array}{l}\sf b)\\\rm x^2+6x>0\\\underline{\sf fac_{\!\!.}a}\\\rm f(x)=x^2+6x\\\underline{\sf zeros\,de\,f(x):}\\\rm x^2+6x=0\\\rm x\cdot(x+6)=0\\\rm x=0\\\rm x+6=0\\\rm x=-6\\\underline{\sf Estudo\,do\,sinal:}\\\rm f(x)>0\longrightarrow x<-6\,ou\,x>0\end{array}}$

$\large\boxed{\begin{array}{l}\rm A\,soluc_{\!\!,}\tilde ao\,\acute e\,dada\,por\\\rm S=\{x\in\mathbb{R}/x<-6\,ou\,x>0\}\end{array}}$

$\large\boxed{\begin{array}{l}\sf c)\\\rm -x^2-4x-4<0\\\underline{\sf fac_{\!\!,}a}\\\rm g(x)=-x^2-4x-4\\\underline{\sf zeros\,de\,g(x):}\\\rm -x^2-4x-4=0\cdot(-1)\\\rm x^2+4x+4=0\\\rm (x+2)^2=0\\\rm x+2=0\\\rm x=-2\\\underline{\sf estudo\,do\,sinal}\\\rm f(x)<0\,\forall\,x\in\mathbb{R},x\ne-2\\\rm portanto\\\rm S=\{x\in\mathbb{R}/x\ne-2\}\end{array}}$