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

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