Question

Calculați: (0.4)^-2 • (125/8)^-2/3.

Answer 1

Răspuns:

(0.4=4/10

(4/10)⁻²=(10/4)²=100/16

(125/8)^-2/3=(8/125)^2/3=

∛(8/125)²=(2/5)²=4/25

100/16*4/25=100/25*14/16=4·1/4=1

Explicație pas cu pas:

Answer 2

$\it (0,4)^{-2}\cdot\Big(\dfrac{125}{8}\Big)^{-\dfrac{2}{3}}=$

$\it 0,4^{-2}\cdot\Big(\dfrac{5^{3}}{2^{3}}\Big)^{-\dfrac{2}{3}}=$

$\it 0,4^{-2}\cdot\Big(\Big(\dfrac{5}{2}\Big)^{3}\Big)^{-\dfrac{2}{3}}=$

$\it 0,4^{-2}\cdot\Big(\dfrac{5}{2}\Big)^{3\cdot\dfrac{-2}{3}}=$

$\it 0,4^{-2}\cdot\Big(\dfrac{5}{2}\Big)^{\not3\cdot\dfrac{-2}{\not3}}=$

$\it 0,4^{-2}\cdot\Big(\dfrac{5}{2}\Big)^{-2}=$

$\it \Big(0,4\cdot\dfrac{5}{2}\Big)^{-2}=$

$\it \Big(\dfrac{4}{10} \cdot\dfrac{5}{2}\Big)^{-2}=$

$\it \Big(\dfrac{\not4}{\not10} \cdot\dfrac{5}{2}\Big)^{-2}=$

$\it \Big(\dfrac{2}{5} \cdot\dfrac{5}{2}\Big)^{-2}=$

$\it \Big(\dfrac{\not2}{\not5} \cdot\dfrac{\not5}{\not2}\Big)^{-2}=$

$\it 1 ^{-2}=$

$\boxed{\it 1 }$

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