Question

oblicz zadania, (iloczyn i iloraz poteg o jednakowych podstawach) ​​

Answer 1

Odpowiedź:

1.

a) 2^6*2^4=2^(6+4)=2^10

b) (1/3)^4*(1/3)^7=(1/3)^(4+7)=(1/3)^11

c) (-2)^3*(-2)^4=(-2)^(3+4)=(-2)^7

d) (0,24)^5*(0,24)^2=(0,24)^7

e) 3^6*3^3^2=3^(6+1+2)=3^9

f) (2 1/4)^3*(2 1/4)^2*(2 1/4)^5=(2 1/4)^(3+2+5)=(2 1/4)^10

g) (-3)^3*(-3)^4*(-3)^5=(-3)^(3+4+5)=(-3)^12

h) (1,3)^6*(1,3)^4*(1,3)^2=(1,3)^(6+4+2)=(1,3)^12

i) (0,5)^4*(0,5)^0*(0,5)^6=(0,5)^(4+0+6)=(0,5)^10

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2.

a) 12^10:12^4=12^(10-4)=12^6

b) (0,4)^15:(0,4)^9=(0,4)^(15-9)=(0,4)^6

c) (-7)^13:(-7)^8=(-7)^(13-8)=(-7)^5

d) (-1 1/2)^7:(-1 1/2)^5=(-1 1/2)^(7-5)=(-1 1/2)^2

e) 15^8/15=15^(8-1)=15^7

f) 10^16:10^0:10^2:10=10^(16-0-2-1)=10^13

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3.

a) 2^8*2^4 : 2^3=2^(8+4):2^3=2^12:2^3=2^(12-3)=2^9

b) (5^6:5^4*5^7) : (5*5^4)=5^(6-4+7):5^(1+4)=5^(9-5)=5^4

c) (-3)^9:(-3)^3*(-3)^2 : (-3)^2*(-3)^0=(-3)^(9-3+2):(-3)^(2+0)=(-3)^(8-2)=(-3)^6

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4.

a) (-10)^5*(-10)^0*(-10)=(-10)^(5+0+1)=(-10)^6=1 000 000

b) (2 1/4)^3*(2 1/4)^2:(2 1/4)^5=(2 1/4)^(3+2-5)=(2 1/4)^0=1  

c) (5^3*5^2:5^5)*(5^4:5^2)=5^(3+2-5)*5^(4-2)=5^(0+2)=5^2=25

d) (-1/2)^4*(-1/2)^3:(-1/2)^5*(-1/2)^3=(-1/2)^(4+3):(-1/2)^(5+3)=(-1/2)^(7-8)=(-1/2)^(-1)=-2

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Szczegółowe wyjaśnienie: