Question

repede plsss.dau coroana si ma abonez​

Answer 1

Răspuns: Ai demonstrația mai jos

Explicație pas cu pas:

Salutare!

$\bf \{[(3^{n+2} +3^{n+3} -4\cdot 3^{n+1}):(2\cdot3^{n+3} -5\cdot 3^{n+2} -3^{n})]^{2}\}^{100}\cdot (9^{638} )^{3}=$

$\bf \{[(3^{n+1}\cdot(3^{1} +3^{2} -4\cdot 3^{0})):(3^{n}\cdot(2\cdot3^{3} -5\cdot 3^{2} -3^{0}))]^{2}\}^{100}\cdot (9^{638} )^{3}=$

$\bf \{[(3^{n+1}\cdot(3 +9-4\cdot 1)):(3^{n}\cdot(2\cdot27 -5\cdot9 -1))]^{2}\}^{100}\cdot (9^{638} )^{3}=$

$\bf \{[(3^{n+1}\cdot 8) :3^{n}\cdot(54 -45 -1)]^{2}\}^{100}\cdot (9^{638} )^{3}=$

$\bf \{[(3^{n+1}\cdot 8) :(3^{n}\cdot 8)]^{2}\}^{100}\cdot (9^{638} )^{3}=$

$\bf [(3^{n+1-n})^{2}]^{100}\cdot (9^{638} )^{3}=$

$\bf [(3^{1})^{2}]^{100}\cdot (9^{638} )^{3}=$

$\bf (3^{1\cdot2})^{100}\cdot 9^{638\cdot 3}=$

$\bf (3^{2})^{100}\cdot 9^{1914}=$

$\bf 9^{100}\cdot 9^{1914}=$

$\bf 9^{100+1914}=$

$\boxed{\bf 9^{2014}}$

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  

==pav38==