Question

Seja x um número real, determine o conjunto solução da seguinte inequação exponencial:
${2}^{2x + 2} { - 2}^{x + 3} > {2}^{x} - 2$
ME AJUDEM!!​

Answer 1

Resposta:

$S=\{x\in IR/x<-2~~oux>1\}$

Explicação passo-a-passo:

$2^2^x^+^2-2^x^+^3>2^x-2\\\\2^2^x.2^2-2^x.2^3>\frac{2^x}{2} \\\\(2^x)^2.4-2^x.8>2^x-2\\\\2^x=y \\\\y^2.4-y.8>y-2\\\\4y^2-8y-y+2>0\\\\4y^2-9y+2>0\\\\Ra \acute ises\\\\4y^2-9y+2=0\\\\\Delta=9^2-4.4.2\\\\81-32\\\\\Delta=49\\\\y=\frac{-(-9)\pm\sqrt{49} }{2.4}\\\\y=\frac{9\pm7}{8}\\\\y=\frac{16}{8}=2\\ou\\y=\frac{2}{8}=\frac{1}{4}\\ \\y=2^_^-2$

_____________2_^-2_______________2____________

              +                         -                                     +

$y<2^-^2 ~~ou~y>2$

$2^x<2^-^2\\\\x<-2\\\\ou\\\\2^x>2\\\\x>1$