Ответ: x∈(-∞;1).
Объяснение:
(x³+2*2ˣ+2)³>(x³+4ˣ+2ˣ)³
∛(x³+2*2ˣ+2)³>∛(x³+4ˣ+2ˣ)³
x³+2*2ˣ+2>x³+4ˣ+2ˣ
2*2ˣ+2>4ˣ+2ˣ
2²ˣ+2ˣ-2*2ˣ-2<0
2²ˣ-2ˣ-2<0
Пусть 2ˣ=t ⇒
t²-t-2<0
t²-t-2=0 D=9 √D=3
t₁=-1 t₂=2
(t+1)(t-2)=0 ⇒
(2ˣ+1)(2ˣ-2)<0
2ˣ+1>0 ⇒
2ˣ-2<0
2ˣ<2¹
x<1.
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Verified answer
Ответ: x∈(-∞;1).
Объяснение:
(x³+2*2ˣ+2)³>(x³+4ˣ+2ˣ)³
∛(x³+2*2ˣ+2)³>∛(x³+4ˣ+2ˣ)³
x³+2*2ˣ+2>x³+4ˣ+2ˣ
2*2ˣ+2>4ˣ+2ˣ
2²ˣ+2ˣ-2*2ˣ-2<0
2²ˣ-2ˣ-2<0
Пусть 2ˣ=t ⇒
t²-t-2<0
t²-t-2=0 D=9 √D=3
t₁=-1 t₂=2
(t+1)(t-2)=0 ⇒
(2ˣ+1)(2ˣ-2)<0
2ˣ+1>0 ⇒
2ˣ-2<0
2ˣ<2¹
x<1.