Question

Calculati [2+2² ×2²²+2²⁰⁰:2¹¹⁹+(3³)¹⁰]:[1 +2²³+2⁸⁰+3³⁰]-2+(5⁷)³=
25 DE PUNCTE + COROANĂ! URGENT! VA ROG!​

Answer 1

Explicație pas cu pas:

Salutare!

$\bf (2 + {2}^{2}\cdot {2}^{22} + {2}^{200} : {2}^{119} + 2\cdot( {3}^{3})^{10}): (1 + {2}^{23} + {2}^{80} + {3}^{30}) - 2+( {5}^{7} )^{3} =$

$\bf (2 + {2}^{2 + 22} + {2}^{200 - 119} + 2\cdot{3}^{3 \cdot10}): (1 + {2}^{23} + {2}^{80} + {3}^{30}) - 2 + {5}^{7 \cdot 3} =$

$\bf (2 + {2}^{24} + {2}^{81} + 2\cdot{3}^{30}) : (1 + {2}^{23} + {2}^{80} + {3}^{30})- 2 + {5}^{21} =$

$\bf 2 \cdot(1 + {2}^{24 - 1} + {2}^{81 - 1} + 1\cdot{3}^{30}) : (1 + {2}^{23} + {2}^{80} + {3}^{30})- 2 + {5}^{21} =$

$\bf 2 \cdot(1 + {2}^{23} + {2}^{80} + {3}^{30}) : (1 + {2}^{23} + {2}^{80} + {3}^{30})- 2 + {5}^{21} =$

$\bf 2 - 2 + {5}^{21} =$

$\boxed{\bf {5}^{21}}$

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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