Question

Prosze o szybką pomoc! Matma. Dam naj. Zadanie nr.8 w załączniku

Answer 1

Odpowiedź:

zad 8

a)

8⁻² = (1/8)² = 1/64

(- 8)² = 64

- 8⁻² = - (1/8)² = - 1/64

- (- 1/2)² = - 1/4

b)

4³ = 64

(- 4)⁻³ = (- 1/4)³ = - 1/64

- 4⁻³ = - (1/4)³ = - 1/64

- (- 1/4)⁻³ = - (- 4)³ = - (- 64) = 64

c)

(1/3)⁻⁴ = 3⁴ = 81

(- 1/3)⁴ = (1/3)⁴ = 1/81

- (- 1/3)⁻⁴ = - (- 3)⁴ = - 81

d)

(√2)⁻⁶ = (1/√2)⁶ = 1/(2¹⁾²)⁶ =1/2⁶⁾² = 1/2³ = 1/8

(- √2)⁶ = (√2)⁶ = (2¹⁾²)⁶ = 2⁶⁾³ = 2³ = 8

- (- √2)⁻⁶ = - (- 1/√2)⁶ = - (1/√2)⁶ = - 1/8

¹⁾² - znaczy do potęgi 1/2