Question

2013•2012+2013•2011-2013•2022+6•[7+3•(5 la puterea2+49 la puterea2 plizzzz repede

Answer 1

Răspuns: 4 071 723

Explicație pas cu pas:

Salutare!

$\bf 2013\cdot2012+2013\cdot2011-2013\cdot2022+6\cdot[7+3\cdot(5^{2} +49^{2})]=$

$\bf 2013\cdot(2012+2011-2022)+6\cdot[7+3\cdot(25 +49^{2})]=$

$\bf 2013\cdot(4023-2022)+6\cdot(7+75 +3\cdot49^{2})=$

$\bf 2013\cdot 2001 +6\cdot(82 +3\cdot49^{2})=$

$\bf 2013\cdot 2001 +492 +18\cdot49^{2}=$

$\bf 2013\cdot 2001 +492 +18\cdot 2401=$

$\bf 2013\cdot 2001 +492 +43218=$

$\bf 4028013 +492 +43218=$

$\boxed{\bf 4071723}$

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  

==pav38==

Answer 2

Răspuns:

4071723

Explicație pas cu pas:

2013×2012+2013×2011-2013×2022+6×[7+3×(5²+49²)]=4050156+4048143-4070286+6[7+3(25+49²)]=4050156+4048143-4070286+6(7+75+3×49²)=4050156+4048143-4070286+492+18×49²=4028505+18×2401=4028505+43218=4071723