Question

(-4/3) la puterea 22:((4/3×(4/3)la puterea 3)totul la puterea 5-(-5/6) la puterea 2
Va rog este urgent!!​

Answer 1

$\Big( - \dfrac{4}{3} \Big)^{22}\div \Big(\dfrac{4}{3}\cdot \Big(\dfrac{4}{3} \Big)^{3} \Big)^{5} - \Big( - \dfrac{5}{6} \Big)^{2}$

$\Big( \dfrac{4}{3} \Big)^{22}\div \Big( \Big(\dfrac{4}{3} \Big)^{1 + 3} \Big)^{5} - \dfrac{5^{2} }{6^{2} }$

$\Big( \dfrac{4}{3} \Big)^{22}\div \Big(\dfrac{4}{3} \Big)^{4 \cdot5} - \dfrac{25 }{36}$

$\Big( \dfrac{4}{3} \Big)^{22}\div \Big(\dfrac{4}{3} \Big)^{20} - \dfrac{25 }{36}$

$\Big( \dfrac{4}{3} \Big)^{22 - 20}- \dfrac{25 }{36}$

$\Big( \dfrac{4}{3} \Big)^{2}- \dfrac{25 }{36}$

$\dfrac{4^{2} }{3^{2} }- \dfrac{25 }{36}$

$\dfrac{16}{9} - \dfrac{25 }{36}$

$\dfrac{64-25}{36}$

$\dfrac{39}{36}$

$\dfrac{\not 39}{\not 36}$

$\boxed{\dfrac{13}{12}}$

Cateva formule pentru puteri

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

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

• 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

• (aⁿ)ᵇ = aⁿ ˣ ᵇ sau aⁿ ˣ ᵇ = (aⁿ) ᵇ

• a⁰ = 1 sau 1 = a⁰

Answer 2

Răspuns:

Explicație pas cu pas:

(-4/3)^22:[(-4/3)^(1+3)]^5-(-5/6)²=(-4/3)^22:(-4/3)^20-25/36=

=(-4/3)²-25/36=16/9-25/36=(4/3+5/6)(4/3-5/6)=13/6×3/6=13/12