Question

34. calculati pe operatii;
(3²+4²) ÷ 5²+ (5¹⁰) ²÷ 5¹⁷ + (2³)² - 10001⁰ + 1²⁰ =??

( 8 ⁶)¹²÷ 8 ⁷⁰ - ( 2² + 3²) · 101⁰ + 1³¹=???
URGENTTTT VA ROGGG!!!!!!!!!!!!
DAU TOATE 59 DE PCTE.

Answer 1

Salutare!!

$\bf(3 ^{2} + 4 ^{2}) : 5^{2} + (5 ^{10} )^{2}: 5^{17} + ( {2}^{3})^{2} - 10001^{0} + {1}^{20} =$

$\bf(9+ 16): 5^{2} + 5 ^{10 \cdot2} : 5^{17} + {2}^{3 \cdot 2} - 1 + 1 =$

$\bf 25 : 5^{2} + 5 ^{20} : 5^{17} + {2}^{6} - 1 + 1 =$

$\bf {5}^{2} : 5^{2} + 5 ^{20 - 17} + {2}^{6} - 1 + 1 =$

$\bf {5}^{2 - 2} + 5 ^{3} + {2}^{6} - 1 + 1 =$

$\bf {5}^{0} + 5 ^{3} + {2}^{6} =$

$\bf 1 + 125 + 64 =$

$\boxed{\bf 190}$

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$\bf(8 ^{6}) ^{12}: 8^{70} - (2^{2} + {3}^{2}) \cdot 101^{0} + 1^{31} =$

$\bf 8 ^{6 \cdot12} :8^{70} - (4 + 9) \cdot 1+ 1 =$

$\bf 8 ^{72} : 8^{70} - 13 \cdot 1+ 1 =$

$\bf 8 ^{72 - 70} - 13 \cdot 1+ 1 =$

$\bf 8 ^{2} - 13 + 1 =$

$\bf 64- 13 + 1 =$

$\boxed{\bf 52}$

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==