Question

Rezultate la toate exercițiile va rog Dau coroana

Answer 1

a)

$\bf (25^{4})^{3}:(4^{2}+3^{2})^{11}-(0^{2010}+1^{2012}+2013^{0})=$

$\bf 25^{4\cdot3}:(16+9)^{11}-(0+1+1)=$

$\bf 25^{12}:25^{11}-2=$

$\bf 25^{12-11}-2=$

$\bf 25^{1}-2=$

$\bf 25-2=$

$\boxed{\bf 23}$

b)

$\bf [123 +2^{7} \cdot 2^{2} +2^{67}:2^{27}]:[123 +2^{14}:2^{5}+(2^{5})^{8}}]=$

$\bf [123 +2^{7+2}+2^{67-27}]:[123 +2^{14-5}+2^{5\cdot8}}]=$

$\bf [123 +2^{9}+2^{40}]:[123 +2^{9}+2^{40}}]=$

$\boxed{\bf 1}$

c)

$\bf (2^{8}+ 9^{2}+5^{6}+36^{2}}):(0^{6}+4^{4}+3^{4}+25^{3}+6^{4})=$

$\bf (2^{8}+ 81+5^{6}+(6^{2})^{2}}):(0+(2^{2})^{4}+81+(5^{2})^{3}+6^{4})=$

$\bf (2^{8}+ 81+5^{6}+6^{2\cdot2}}):(2^{2\cdot4}+81+5^{2\cdot3}+6^{4})=$

$\bf (2^{8}+ 81+5^{6}+6^{4}}):(2^{8}+81+5^{6}+6^{4})=$

$\boxed{\bf 1}$

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

(- a)ⁿ,unde n este o putere impara (-a)ⁿ=(-a)ⁿ

(- a)ⁿ,unde n este o putere para (-a)ⁿ = aⁿ