Resposta:
[tex]\textsf{Leia abaixo}[/tex]
Explicação passo a passo:
[tex]\sf{16 < \left(\dfrac{1}{4}\right)^{log_{\frac{1}{5}}\:(x^2 - x + 19)}[/tex]
[tex]\sf{16 < \left(\dfrac{1}{4}\right)^{-log_{5}\:(x^2 - x + 19)}[/tex]
[tex]\sf{\left(\dfrac{1}{4}\right)^{-log_{5}\:(x^2 - x + 19)} > 16}[/tex]
[tex]\sf{4^{log_{5}\:(x^2 - x + 19)} > 4^2}[/tex]
[tex]\sf{log_{5}\:(x^2 - x + 19) > 2}[/tex]
[tex]\sf{log_{5}\:(x^2 - x + 19) > log_5\:5^2}[/tex]
[tex]\sf{x^2 - x + 19 > 25}[/tex]
[tex]\sf{x^2 - x - 6 > 0}[/tex]
[tex]\mathsf{\Delta = b^2 - 4.a.c}[/tex]
[tex]\mathsf{\Delta = (-1)^2 - 4.1.(-6)}[/tex]
[tex]\mathsf{\Delta = 1 + 24}[/tex]
[tex]\mathsf{\Delta = 25}[/tex]
[tex]\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{1 \pm \sqrt{25}}{2} \rightarrow \begin{cases}\mathsf{x' = \dfrac{1 + 5}{2} = \dfrac{6}{2} = 3}\\\\\mathsf{x'' = \dfrac{1 - 5}{2} = -\dfrac{4}{2} = -2}\end{cases}}[/tex]
[tex]\boxed{\boxed{\mathsf{S = \{\:\:\:]-\infty,-2\:[ \:\:\:\cup\:\:\: ]\:3,+\infty[\:\:\:\}}}}[/tex]
