Question

+-12≤0

Answer 1

Ответ: x∈(-∞;1].

Пошаговое объяснение:

3²ˣ+3ˣ-12≤0

Пусть 3ˣ=t>0   ⇒

t²+t-12≤0

t²+t-12=0   D=49      √D=7

t₁=3      t₂=-4

(t-3)*(t+4)≤0

-∞__+__-4__-__3__+__+∞

t∈[-4;3]

Так как t>0   ⇒

t∈(0;3]

3ˣ∈(0;3]   ⇒

x∈(-∞;1].

Answer 2

Ответ: х ∈ (-∞; 1)

(фото)

Answer 3

This inequality is true because both 12 and -12 are less than or equal to 0. So,

+12 ≤ 0 and

-12 ≤ 0

Therefore, the inequality +-12 ≤ 0 is true.