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kotuk2000
@kotuk2000
July 2022
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Подробное решение, пожалуйста
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oganesbagoyan
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Подробное решение, пожалуйста
( ∛(6 +√35) ) ^x + (
∛(6 -√35) ) ^x =12 ;
( (∛(6+√35) ) ^x ) * ( ( ∛(6 -√35) ) ^x) =1 * * * обратные числа * * *
---
замена : t = (
∛(6 +√35) ) ^x ≠0 ⇒ (
∛(6 -√35) ) ^x =1 /t , получаем :
t +1/t =12 ;
t
² -12t +1 =0 , D/4 =6² -1 =35 =(√35)²
t₁ = 6+ √35 ;
t₂ =6 - √35 = 1/(6+ √35) = (6+ √35)⁻
¹ .
---
(∛(6 +√35) ) ^x = t₁ ;
(∛(6 +√35) ) ^x = 6+ √35 ⇔( 6+ √35) ^ (x/3) = (6+ √35) ⇒ x/3 =1 ⇔ x=3.
---
(∛(6 +√35) ) ^x =
t₂ ;
(∛(6 +√35) ) ^x =( 6+ √35)⁻¹ ⇔( 6+ √35) ^ (x/3) = (6+ √35)⁻¹ ⇒ x/3 = -1⇔ x=-3.
ответ : ± 3.
* * * * * * *
* * * * * * *
∛(6+√35) ) ^x * ( ∛(6 -√35) ) ^x = ( ∛(6 +√35) *
∛(6 -√35) ) ^x =
(∛ (6 +√35) *(6 -√35) ) ^x = (∛ (6² -(√35)²) ) ^x = (∛ (36 -35) ) ^x = 1.
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Answers & Comments
Verified answer
Подробное решение, пожалуйста( ∛(6 +√35) ) ^x + ( ∛(6 -√35) ) ^x =12 ;
( (∛(6+√35) ) ^x ) * ( ( ∛(6 -√35) ) ^x) =1 * * * обратные числа * * *
---
замена : t = (∛(6 +√35) ) ^x ≠0 ⇒ ( ∛(6 -√35) ) ^x =1 /t , получаем :
t +1/t =12 ;
t² -12t +1 =0 , D/4 =6² -1 =35 =(√35)²
t₁ = 6+ √35 ;
t₂ =6 - √35 = 1/(6+ √35) = (6+ √35)⁻¹ .
---
(∛(6 +√35) ) ^x = t₁ ;
(∛(6 +√35) ) ^x = 6+ √35 ⇔( 6+ √35) ^ (x/3) = (6+ √35) ⇒ x/3 =1 ⇔ x=3.
---
(∛(6 +√35) ) ^x =t₂ ;
(∛(6 +√35) ) ^x =( 6+ √35)⁻¹ ⇔( 6+ √35) ^ (x/3) = (6+ √35)⁻¹ ⇒ x/3 = -1⇔ x=-3.
ответ : ± 3.
* * * * * * * * * * * * * *
∛(6+√35) ) ^x * ( ∛(6 -√35) ) ^x = ( ∛(6 +√35) * ∛(6 -√35) ) ^x =
(∛ (6 +√35) *(6 -√35) ) ^x = (∛ (6² -(√35)²) ) ^x = (∛ (36 -35) ) ^x = 1.