Szczegółowe wyjaśnienie:
6.6.
[tex]a)\ 1^3=1;\ (-1)^2=1;\ 1^2=1;\ (-1)^5=-1\\\\b)\ 2^0=1;\ 2^2=2;\ (-2)^3=-8;\ 2^5=32;\ 2^6=64;\ (-2)^4=16\\\\c)\ \left(-\dfrac{2}{5}\right)^2=\dfrac{4}{25};\ \left(-\dfrac{2}{3}\right)^1=-\dfrac{2}{3};\ \left(\dfrac{3}{2}\right)^3=\dfrac{3^3}{2^3}=\dfrac{27}{8}\\\\d)\ 0,1^2=0,01;\ 0,2^3=0,008;\ 0,4^2=0,16;\ 0,05^1=0,05[/tex]
6.7.
Zastosujemy twierdzenie:
[tex]a^n\cdot a^m=a^{n+m}[/tex]
[tex]a) 2^3\cdot2^2=2^5\\\\b)\ 2^1\cdot2^2\cdot2^3=2^6\\\\c)\ 3^2\cdot3^3\cdot3^0=3^5\\\\d)\ \left(\dfrac{1}{3}\right)\cdot\left(\dfrac{1}{3}\right)^2\cdot\left(\dfrac{1}{3}\right)^6=\left(\dfrac{1}{3}\right)^9\\\\e)\ (-2)^2\cdot(-2)^3\cdot(-2)^4=(-2)^9\\\\f)\ 10\cdot10^2\cdot10^3=10^6\\\\g)\ (-10)\cdot(-10)^3\cdot(-10)^5=(-10)^9\\\\h)\ 0,2\cdot0,2^2\cdot0,2^3\cdot0,2^4=0,2^{10}[/tex]
