Salutare!!
[tex]\bf(3 ^{2} + 4 ^{2}) : 5^{2} + (5 ^{10} )^{2}: 5^{17} + ( {2}^{3})^{2} - 10001^{0} + {1}^{20} =[/tex]
[tex]\bf(9+ 16): 5^{2} + 5 ^{10 \cdot2} : 5^{17} + {2}^{3 \cdot 2} - 1 + 1 =[/tex]
[tex]\bf 25 : 5^{2} + 5 ^{20} : 5^{17} + {2}^{6} - 1 + 1 =[/tex]
[tex]\bf {5}^{2} : 5^{2} + 5 ^{20 - 17} + {2}^{6} - 1 + 1 =[/tex]
[tex]\bf {5}^{2 - 2} + 5 ^{3} + {2}^{6} - 1 + 1 = [/tex]
[tex]\bf {5}^{0} + 5 ^{3} + {2}^{6} = [/tex]
[tex]\bf 1 + 125 + 64 = [/tex]
[tex] \boxed{\bf 190}[/tex]
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[tex]\bf(8 ^{6}) ^{12}: 8^{70} - (2^{2} + {3}^{2}) \cdot 101^{0} + 1^{31} =[/tex]
[tex]\bf 8 ^{6 \cdot12} :8^{70} - (4 + 9) \cdot 1+ 1 =[/tex]
[tex]\bf 8 ^{72} : 8^{70} - 13 \cdot 1+ 1 =[/tex]
[tex]\bf 8 ^{72 - 70} - 13 \cdot 1+ 1 =[/tex]
[tex]\bf 8 ^{2} - 13 + 1 =[/tex]
[tex]\bf 64- 13 + 1 =[/tex]
[tex] \boxed{\bf 52}[/tex]
Cateva formule pentru puteri
a⁰ = 1 sau 1 = a⁰
1 ⁿ = 1 sau a¹ = a
(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
(- a)ⁿ,unde n este o putere impara (-a)ⁿ=(-a)ⁿ
(- a)ⁿ,unde n este o putere para (-a)ⁿ = aⁿ
==pav38==
