Question

pls va rog repede, mulțumesc!​

Answer 1

Răspuns:

• (B) n = 1006
• (C) n = 55
• (D) n = 10

Explicație pas cu pas:

Salutare!

(B)

$\bf 9^{n}+9^{n+1}= 10\cdot 3^{2012}$

$\bf 9^{n}\cdot( 9^{n-n}+9^{n+1-n})= 10\cdot 3^{2012}$

$\bf (3^{2})^{n}\cdot( 9^{0}+9^{1})= 10\cdot 3^{2012}$

$\bf 3^{2\cdot n}\cdot 10= 10\cdot 3^{2012}\:\:\:\Big|:10$

$\bf 3^{2n} = 3^{2012}\implies2n=2012\:\:\Big|:2\implies\boxed{\bf n = 1006}$

(C)

$\bf 6^{n}+6^{n+3}= 217\cdot 6^{55}$

$\bf 6^{n}\cdot( 6^{n-n}+6^{n+3-n})= 217\cdot 6^{55}$

$\bf 6^{n}\cdot( 6^{0}+6^{3})= 217\cdot 6^{55}$

$\bf 6^{n}\cdot( 1+2016)= 217\cdot 6^{55}$

$\bf 6^{n}\cdot 217= 217\cdot 6^{55}\:\:\:\Big|:217$

$\bf 6^{n}=6^{55}\implies \boxed{\bf n = 55}$

(D)

$\bf 7^{n+1}+7^{n+2}= 8\cdot 7^{11}$

$\bf 7^{n+1}\cdot(7^{n+1-(n+1)}+7^{n+2-(n+1)})= 8\cdot 7^{11}$

$\bf 7^{n+1}\cdot(7^{0}+7^{1})= 8\cdot 7^{11}$

$\bf 7^{n+1}\cdot(1+7)= 8\cdot 7^{11}$

$\bf 7^{n+1}\cdot 8 = 8\cdot 7^{11}\:\:\:\Big|:8$

$\bf 7^{n+1} = 7^{11} \implies n+1 = 11 \implies n=11-1 \implies \boxed{\bf n = 10}$

Formule pentru puteri

a⁰ = 1 sau 1 = a⁰

(aⁿ)ᵇ = aⁿ ˣ ᵇ sau aⁿ ˣ ᵇ = (aⁿ) ᵇ

aⁿ · aᵇ = (a · a) ⁿ ⁺ ᵇ  sau  (a · a) ⁿ ⁺ ᵇ = aⁿ · aᵇ

aⁿ : aᵇ = (a : a) ⁿ ⁻ ᵇ sau (a : a) ⁿ ⁻ ᵇ = aⁿ : aᵇ

aⁿ · bⁿ = (a · b)ⁿ sau (a · b)ⁿ = aⁿ · bⁿ

aⁿ : bⁿ = (a : b)ⁿ sau (a : b)ⁿ = aⁿ : bⁿ

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