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(-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!!​


Răspuns :

[tex] \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} [/tex]

[tex]\Big( \dfrac{4}{3} \Big)^{22}\div \Big( \Big(\dfrac{4}{3} \Big)^{1 + 3} \Big)^{5} - \dfrac{5^{2} }{6^{2} }[/tex]

[tex]\Big( \dfrac{4}{3} \Big)^{22}\div \Big(\dfrac{4}{3} \Big)^{4 \cdot5} - \dfrac{25 }{36}[/tex]

[tex]\Big( \dfrac{4}{3} \Big)^{22}\div \Big(\dfrac{4}{3} \Big)^{20} - \dfrac{25 }{36}[/tex]

[tex]\Big( \dfrac{4}{3} \Big)^{22 - 20}- \dfrac{25 }{36}[/tex]

[tex]\Big( \dfrac{4}{3} \Big)^{2}- \dfrac{25 }{36}[/tex]

[tex]\dfrac{4^{2} }{3^{2} }- \dfrac{25 }{36}[/tex]

[tex]\dfrac{16}{9} - \dfrac{25 }{36}[/tex]

[tex]\dfrac{64-25}{36}[/tex]

[tex]\dfrac{39}{36}[/tex]

[tex]\dfrac{\not 39}{\not 36}[/tex]

[tex] \boxed{\dfrac{13}{12}}[/tex]

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

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