Răspuns: Am demonstrat mai jos
Explicație pas cu pas:
Salutare!
[tex]\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}=[/tex]
[tex]\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}=[/tex]
[tex]\bf \{[(3^{n+1}\cdot(3 +9-4\cdot 1)):(3^{n}\cdot(2\cdot27 -5\cdot9 -1))]^{2}\}^{100}\cdot (9^{638} )^{3}=[/tex]
[tex]\bf \{[(3^{n+1}\cdot 8) :3^{n}\cdot(54 -45 -1)]^{2}\}^{100}\cdot (9^{638} )^{3}=[/tex]
[tex]\bf \{[(3^{n+1}\cdot 8) :(3^{n}\cdot 8)]^{2}\}^{100}\cdot (9^{638} )^{3}=[/tex]
[tex]\bf [(3^{n+1-n})^{2}]^{100}\cdot (9^{638} )^{3}=[/tex]
[tex]\bf [(3^{1})^{2}]^{100}\cdot (9^{638} )^{3}=[/tex]
[tex]\bf (3^{1\cdot2})^{100}\cdot 9^{638\cdot 3}=[/tex]
[tex]\bf (3^{2})^{100}\cdot 9^{1914}=[/tex]
[tex]\bf 9^{100}\cdot 9^{1914}=[/tex]
[tex]\bf 9^{100+1914}=[/tex]
[tex]\boxed{\bf 9^{2014}}[/tex]
Am folosit formulele
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
==pav38==
