Czy mógłby ktos robić zad. 3 i 4? :) Będę bardzo wdzięczna.
3.
a) 25^20:25^5=25^15
b) d^50:d^37=d^14
c) 10^-12:10^-12=10^1
d) 7^-9:7^5=7^-14
e) (-3,9)^6:(1,3)^6=3^6
f) 12^9:4^9=3^9
g) (2 1/3)^5: (1 2/5)^5= (7/3)^5:(7/5)^5= (5/3)^5=(1 2/3)^5
4.
a) (5^2 * 5^8 * 5^4) : (5^6 * 5^4* 5^2)= (5^14 : 5^12=5^2=25
b) (-3)^16 : (-3)^8 / (-13)^12 : (-13)^7= (-13)^8 : (-13)^7= (-13)^1=-13
c) (1/4)^100 * 4^100= 1^100= 1
d) (2^2 * 3^2 * 4^2* 5^2) : (3^2 * 8^2 * 10^2)^3= (60)^6 : (180)^6= [0,(3)]^6 ( 0,3 w okresie)
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3.
a) 25^20:25^5=25^15
b) d^50:d^37=d^14
c) 10^-12:10^-12=10^1
d) 7^-9:7^5=7^-14
e) (-3,9)^6:(1,3)^6=3^6
f) 12^9:4^9=3^9
g) (2 1/3)^5: (1 2/5)^5= (7/3)^5:(7/5)^5= (5/3)^5=(1 2/3)^5
4.
a) (5^2 * 5^8 * 5^4) : (5^6 * 5^4* 5^2)= (5^14 : 5^12=5^2=25
b) (-3)^16 : (-3)^8 / (-13)^12 : (-13)^7= (-13)^8 : (-13)^7= (-13)^1=-13
c) (1/4)^100 * 4^100= 1^100= 1
d) (2^2 * 3^2 * 4^2* 5^2) : (3^2 * 8^2 * 10^2)^3= (60)^6 : (180)^6= [0,(3)]^6 ( 0,3 w okresie)