Răspuns :
Explicație pas cu pas:
Salutare!
[tex]\bf (2 + {2}^{2}\cdot {2}^{22} + {2}^{200} : {2}^{119} + 2\cdot( {3}^{3})^{10}): (1 + {2}^{23} + {2}^{80} + {3}^{30}) - 2+( {5}^{7} )^{3} =[/tex]
[tex]\bf (2 + {2}^{2 + 22} + {2}^{200 - 119} + 2\cdot{3}^{3 \cdot10}): (1 + {2}^{23} + {2}^{80} + {3}^{30}) - 2 + {5}^{7 \cdot 3} =[/tex]
[tex]\bf (2 + {2}^{24} + {2}^{81} + 2\cdot{3}^{30}) : (1 + {2}^{23} + {2}^{80} + {3}^{30})- 2 + {5}^{21} =[/tex]
[tex]\bf 2 \cdot(1 + {2}^{24 - 1} + {2}^{81 - 1} + 1\cdot{3}^{30}) : (1 + {2}^{23} + {2}^{80} + {3}^{30})- 2 + {5}^{21} =[/tex]
[tex]\bf 2 \cdot(1 + {2}^{23} + {2}^{80} + {3}^{30}) : (1 + {2}^{23} + {2}^{80} + {3}^{30})- 2 + {5}^{21} =[/tex]
[tex]\bf 2 - 2 + {5}^{21} = [/tex]
[tex] \boxed{\bf {5}^{21}}[/tex]
Câteva 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
#copaceibrainly
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