Question

Calculati: 2 la puterea 30 : 2 la puterea 25 + 2 la puterea 4 la puterea a - 2 × [ 128:2 la puterea 5 + 2 la puterea 3 : (2×2 la puterea 4 - 24)] ​

Answer 1

Răspuns:

38

Explicație pas cu pas:

2^30:2^25+2^4-2×[128:2^5+2^3 :(2×2^4-24)]=

2^(30-25)+2^4-2×[2^7:2^5+2^3 :(2^5-24)]=

2^5+2^4-2×[2^(7-5)+2^3 :(32 - 24)] =

2^5+2^4-2×(2^2 +2^3 :8) =

2^5+2^4-2×(2^2 +2^3 :2^3) =

2^5+2^4-2×(4 +1) =

32 +16 - 2 ×5 =

32 +16 - 10 =

48 - 10 =

38

Answer 2

Salutare!

$\bf 2^{30}:2^{25}+2^{4^{0}}-2\cdot [128:2^{5}+2^{3} :(2\cdot 2^{4}-24)]=$

$\bf 2^{30-25}+2^{1}-2\cdot [2^{7}:2^{5}+2^{3} :(2^{1+4}-24)]=$

$\bf 2^{5}+2-2\cdot [2^{7-5}+2^{3} :(2^{5}-24)]=$

$\bf 2^{5}+2-2\cdot [2^{2}+2^{3} :(32-24)]=$

$\bf 2^{5}+2-2\cdot (2^{2}+2^{3} :8)=$

$\bf 2^{5}+2-2\cdot (2^{2}+2^{3} :2^{3})=$

$\bf 2^{5}+2-2\cdot (2^{2}+2^{3-3})=$

$\bf 2^{5}+2-2\cdot (4+2^{0})=$

$\bf 2^{5}+2-2\cdot (4+1)=$

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

$\bf 32+2-10=$

$\boxed{\bf 24}$

$\bf 2^{4^{0}} = 2^{1} = 2$

Cateva 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