Question

Помогите,пожалуйстаааа!!!!!!
1/7^(x^2-9)<=1

Answer 1

$(frac{1}{7} )^{x^{2} -9} leq 1$
 $(x^{2} -9) geq 0$
$(x-3)(x+3) geq 0$
x ∈(-00;-3) U (3; +00)

Answer 2

$frac{1}{7^{ x^{2} - 9} } leq 1 \ 1 leq 7^{ x^{2} -9} \ x^{2} - 9 geq log_{7}1 \ x^{2} - 9 geq 0$
x ∈ (-00, -3] ∨ [3, +00)