Question

Resolva as inequação (2/3)^3x-2 . (4/9)^2x+1 menor ou igual que (8/27)^x-3

me ajudem por favor

Answer 1

Resposta:

.      S  =  { x  ∈  R  /  x   ≤   - 9/4 }

Explicação passo a passo:

.

.      (2/3)^3x-2  .  (4/9)^2x+1   ≤   (8/27)^x-3

.      (2/3)^3x-2  .  [(2/3)²]^2x+1   ≤  [(2/3)^3]^x-3

.      (2/3)^3x-2  .  (2/3)^4x+2   ≤   (2/3)^3x-9

.      (2/3)^(3x-2+4x+2)   ≤   (2/3)^3x-9

.      (2/3)^7x   ≤  (2/3)^3x-9                 ( elimina a base (2/3) )

==>  7x   ≤  3x  -  9

.       7x  -  3x   ≤   - 9

.       4x   ≤   - 9

.       x   ≤   - 9/4

.

(Espero ter colaborado)

Answer 2

Explicação passo-a-passo:

$( \frac{2}{3} ) {}^{3x - 2} .( \frac{4}{9} ) {}^{2x +1 } \leqslant ( \frac{8}{27} ) {}^{x - 3}$

$( \frac{2}{3} ) {}^{3x - 2} .( \frac{2}{3} ) {}^{2.(2x + 1)} \leqslant ( \frac{2}{3} ) {}^{3.(x - 3)}$

$( \frac{2}{3} ) {}^{3x - 2 + 4x + 2} \leqslant ( \frac{2}{3} ) {}^{3x - 9}$

$7x \leqslant 3x - 9$

$7x - 3x \leqslant - 9$

$4x \leqslant - 9$

$x \leqslant \frac{ - 9}{4}$