Autorzy:Jerzy Janowicz
Wydawnictwo:Nowa Era
Rok wydania:2016
Oblicz 4.43 gwiazdek na podstawie 7 opinii

`a)\ ((5 1/3)^2*2^-4)/((3/4)^-2)=((5 1/3)^2)/((4/3)^2)*2^-4=(5 1/3:4/3)^2*1/2^4=` 

`\ \ \ =(16/strike3^1*strike3^1/4)^2*1/16=(16/4)^2*1/16=4^2*1/16=16*1/16=1` 

 

`b)\ ((6 1/4)^-2*(2/5)^-5)/(0,5^3)=((25/4)^-2*(5/2)^5)/((1/2)^3)=(((5/2)^2)^(-2)*(5/2)^5)/(1/8)=` 

`\ \ \ =((5/2)^(2*(-2))*(5/2)^5):1/8=((5/2)^-4*(5/2)^5)*8=` 

`\ \ \ =(5/2)^(-4+5)*8=5/2*8=40/2=20` 

 

`c)\ (2,8^-2*7^3)/(0,4^-3)=((0,4*7)^-2*7^3)/(0,4^-3)=(0,4^-2*7^-2*7^3)/(0,4^-3)=(0,4^-2)/(0,4^-3)*7^-2*7^3=` 

`\ \ \ =0,4^(-2-(-3))*7^(-2+3)=0,4^(-2+3)*7=0,4*7=2,8` 

 

`d)\ (9,6^-1*(1/2)^-4)/(1-(3/4)^-2)=(1/(9,6)*2^4)/(1-(4/3)^2)=(10/strike96^(\ 6)*strike16^1)/(1-16/9)=(10/6)/(-7/9)=10/6:(-7/9)=` 

`\ \ \ =5/3:(-7/9)=5/strike3^1*(-strike9^3/7)=-15/7=-2 1/7`