III gimnazjum

Matematyka z plusem 3

Rozwiązanie 1

(−31​)2=(−31​)⋅(−31​)=91​ −0,(7)=−0,777… −0,(70)=−0,7070…

a)23=2⋅2⋅2=4⋅2=8 (53​)2=53​⋅53​=259​ (−1)5=(−1)⋅(−1)⋅(−1)⋅(−1)⋅(−1)=1⋅1⋅(−1)=−1

a)(−3)15=((−1)⋅3)15=(−1)15⋅315=−315&lt;0 (−3)15&lt;314 b)−94=(−1)⋅94&lt;0 (−9)4=((−1)⋅9)4=(−1)4⋅94=1⋅94=94&gt;0

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52x−5=(0,2)3x−5

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8

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Matematyka z plusem 3