Question

ПОМОГИТЕ РЕШИТЬ! !! (2/3)^ (x^2+4x) => (8/27)^ x+2

Answer 1

$(frac{2}{3})^{x^2+4x} geq (frac{8}{27})^{x+2}$
$frac{8}{27}=frac{2^3}{3^3}=(frac{2}{3})^3$
$((frac{2}{3})^3)^{x+2}=(frac{2}{3})^{3(x+2)}=(frac{2}{3})^{3x+6}$
$(frac{2}{3})^{x^2+4x} geq (frac{2}{3})^{3x+6}$
$0<frac{2}{3}<1$
$x^2+4x leq 3x+6$
$x^2+4x-3x-6 leq 0$
$x^2+x-6 leq 0$
$(x+3)(x-2) leq 0$
х є $[-3;2]$