Question

URGENTE!!!!
resolva as inequações exponenciais abaixo:

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\sf{2^{\:x^2-3x}\:\geq\:\dfrac{1}{4}}$

$\sf{2^{\:x^2-3x}\:\geq\:2^{-2}}$

$\sf{x^2 - 3x\:\geq\:-2}$

$\sf{x^2 - 3x + 2\geq 0}$

$\mathsf{\Delta = b^2 - 4.a.c }$

$\mathsf{\Delta = (-3)^2 - 4.1.2 }$

$\mathsf{\Delta = 9 - 8 }$

$\mathsf{\Delta = 1 }$

$\sf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{3 \pm \sqrt{1}}{2} \rightarrow \begin{cases}\sf{x' = \dfrac{3 + 1}{2} = \dfrac{4}{2} = 2}\\\\\sf{x'' = \dfrac{3 - 1}{2} = \dfrac{2}{2} = 1}\end{cases}}$

$\boxed{\boxed{\sf{S = \{\:]-\infty,1] \:\cup\:[2,+\infty[\:\}}}}$