[tex]\large\boxed{\begin{array}{l}(\bf{UF-GO})~\sf Os\,valores\,de\,x\,para\,os\,quais\\\sf (0,8)^{4x^2-x}>(0,8)^{3(x+1)}\\\underline{\sf soluc_{\!\!,}\tilde ao\!:}\\\rm Quando\,a\,base\,est\acute a\,entre\,0\,e\,1\,a\\\rm desigualdade\,se\,inverte.\\\rm (0,8)^{4x^2-x}>(0,8)^{3(x+1)}\\\rm 4x^2-x<3(x+1)\\\rm 4x^2-x<3x+3\\\rm 4x^2-x-3x-3<0\\\rm 4x^2-4x-3<0\end{array}}[/tex]
[tex]\boxed{\begin{array}{l}\underline{\sf fac_{\!\!,}a}\\\rm f(x)=4x^2-4x-3\\\rm a=4>0\,concavidade\,para\,cima\\\underline{\sf zeros\,de\,f(x):}\\\rm 4x^2-4x-3=0\\\rm\Delta=b^2-4ac\\\rm\Delta=(-4)^2-4\cdot4\cdot(-3)\\\rm\Delta=16+48\\\rm\Delta=64\\\rm x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\rm x=\dfrac{-(-4)\pm\sqrt{64}}{2\cdot4}\\\\\rm x=\dfrac{4\pm8}{8}\\\begin{cases}\rm x_1=\dfrac{4+8}{8}=\dfrac{12\div4}{8\div4}=\dfrac{3}{2}\\\rm x_2=\dfrac{4-8}{8}=-\dfrac{4}{8}=-\dfrac{1}{2}\end{cases}\end{array}}[/tex]
[tex]\large\boxed{\begin{array}{l}\underline{\sf estudo\,do\,sinal}\\\rm f(x)>0\longrightarrow x<-\dfrac{1}{2}~ou~x>\dfrac{3}{2}\\\\\rm f(x)<0\longrightarrow -\dfrac{1}{2}<x<\dfrac{3}{2}\\\\\rm queremos\,os\,valores\,de\,x\\\rm para\,os\,quais\, f(x)<0.\\\rm Portanto\,a\,soluc_{\!\!,}\tilde ao\,da\,inequac_{\!\!,}\tilde ao\,\acute e\\\rm S=\bigg\{x\!\in\!\mathbb{R}\bigg/\!\!-\dfrac{1}{2}<x<\dfrac{3}{2}\bigg\}\end{array}}[/tex]
