Question

Ajudem! pfv.

$(3^{-2})^{x+1} \geq 3^{x}$

$( \frac{1}{2})^{4x+1}\ \textless \ (\frac{1}{2})^{3x-1}$

Answer 1

( 3⁻² )ˣ⁺¹ ≥3ˣ

Notas =
( 3⁻² ) = ( 1/3 )²  = 
reescrevendo
[( 1/3)²]ˣ⁺¹ ( 1/3)²ˣ⁺²
reescrevendo
( 1/3)²ˣ⁺²  ≤  ( 1/3)⁻ˣ
2x + 2 ≤ - x
2x + x  ≤  - 2
3x ≤  -2
x≤  - 2/3

2
( 1/2)⁴ˣ ⁺¹ < ( 1/2)³ˣ⁻¹
4x + 1 < 3x - 1
4x - 3x < -1 - 1
x < -2 ***