zad 10 i 11 odrazę wiekie dzięki klasa 6 mnożenie liczb wymiernych ​

Zad 10 I 11 Odrazę Wiekie Dzięki Klasa 6 Mnożenie Liczb Wymiernych class=

Odpowiedź :

Odpowiedź:

Zadanie 10

a) (2/3)^2 = 4/9

(-2/3)^3 = -8/27

(2/3)^5 = 32/243

(-2/3)^6 = 64/729

b) (2 1/6)^2 = (13/6)^2 = 169/36

(-1 1/9)^2 = (-10/9)^2 = 100/81

(2 1/7)^2 = (15/7)^2 = 225/49

(-1 1/11)^2 = (-12/11)^2 = 144/121

(3 1/3)^2 = (10/3)^2 = 100/9

Zadanie 11

a) (-1/10)^2 = 1/100 = 0.01

(-1/10)^3 = -1/1000 = -0.001

(1/10)^4 = 1/10000 = 0.0001

b) (1/10)^5 = 1/100000 = 0.00001

(-1.7)^2 = (17/10)^2 = 289/100 = 2.89

(-1.8)^2 = (18/10)^2 = 324/100 = 3.24

c) 1.2^2 = (12/10)^2 = 144/100 = 1.44

1.3^2 = (13/10)^2 = 169/100 = 1.69

(-1.1)^2 = (11/10)^2 = 121/100 = 1.21