Question

Potęga o wykładniku całkowitym rozwiązać przykłady

Answer 1

Odpowiedź:

zad 1

e) 4⁻⁴ = (1/4)⁴ = 1/256

f) 5⁻³ = (1/5)³ = 1/125

g) (- 6)⁻³ = (- 1/6)³ = - 1/216

h) (- 9)⁻³ = (- 1/9)³ = - 1/729

zad 2

a) (1/4)⁴ = 1/256

b) (2/3)⁴ = 16/81

c) (1 1/2)⁶ = (3/2)⁶ = 729/72 = 10,125

d) (13/7)⁰ = 1

e) (5/9)⁻¹ = 9/5 = 1 4/5

f) (2/3)⁻³ = (3/2)³ = 27/8 = 3 3/8

g) (1 1/4)⁻³ = (5/4)⁻³ = (4/5)³ = 64/125

h) ( 3 1/3)⁻⁴ = (10/3)⁻⁴ = (3/10)⁴ = 81/10000