plis oblicz;) dodałem załocznik mam nadzieje ze mi to wytłumaczysz
Zadanie 1
=
b)
= 3,3*(-27)=-89,1
c)
Zadanie 2
Zadanie 3
a)
Zadanie 4
Zadanie 5
Zadanie 6
d)
Pozdrawiam.
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Zadanie 1
=
=
b)
=
=
=
= 3,3*(-27)=-89,1
c)![(\frac{3^{-2})^3:(3^{-4}*3^{8})^{-2}}{[3^{-2})2]^{-1}} (\frac{3^{-2})^3:(3^{-4}*3^{8})^{-2}}{[3^{-2})2]^{-1}}](https://tex.z-dn.net/?f=%28%5Cfrac%7B3%5E%7B-2%7D%29%5E3%3A%283%5E%7B-4%7D%2A3%5E%7B8%7D%29%5E%7B-2%7D%7D%7B%5B3%5E%7B-2%7D%292%5D%5E%7B-1%7D%7D)
=
=
Zadanie 2
Zadanie 3
a)
b)
Zadanie 4
a)
b)
Zadanie 5
a)
b)
c)![\sqrt[3]{-3 \frac{3}{8}}+ \sqrt[3]-0,125}= \sqrt[3]{-3 \frac{3}{8}}+ \sqrt[3]-0,125}=](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-3+%5Cfrac%7B3%7D%7B8%7D%7D%2B+%5Csqrt%5B3%5D-0%2C125%7D%3D)
=![\sqrt[3]{-\frac{27}{8}}-0,5= \sqrt[3]{-\frac{27}{8}}-0,5=](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-%5Cfrac%7B27%7D%7B8%7D%7D-0%2C5%3D)
=
Zadanie 6
a)
b)
c)
d)![\sqrt[3]{16}=\sqrt[]{8*2}= 2\sqrt[3]{2} \sqrt[3]{16}=\sqrt[]{8*2}= 2\sqrt[3]{2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B16%7D%3D%5Csqrt%5B%5D%7B8%2A2%7D%3D+2%5Csqrt%5B3%5D%7B2%7D)
Pozdrawiam.