Question

 me ajudem a resolver: 
x² ≤ 9

Answer 1

EDITANDO
$x^2 \leq 9\\x' \leq \sqrt{9}\\\\\boxed{x' \leq 3}\\\\x'' \leq -\sqrt{9}\\\\\boxed{x" \leq -3}\\\\\boxed{-3 \leq x \leq 3}$

Answer 2

x²≤9
x≤√9

x≤3
x≥-3

S={x≥-3 ou x≤3}