III gimnazjum

Matematyka wokół nas 3

Obliczmy najpierw wartosˊcˊ tego wyraz˙enia: (32​)3⋅(−121​)3+(8<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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M834 80h400000v40h-400000z'/></svg>​)2:23−(3−4<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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Rozwiązanie 1

a) (2⋅42)2=322=1024 [(2⋅4)2]2=642=4096 `ul(ul((2*4^2)^2

Aby obliczycˊ, jakim ułamkiem liczby a jest liczba b, wystarczy uprosˊcicˊ wyraz˙enie ab​:

(−21​)3=(−2−1)3=−2−3   22=(2−1)−2=(21​)−2

(0,2)2=(51​)2=(5−1)2=5−2 (−31​)3=(−3−1)3=−3−3

a) 4−2=421​=161​ (−12)−3=(−121​)3=−17281​   (−43​)−1=−34​

a)   (231​)−3=(37​)−3=(73​)3=34327​ b)

Wpierw sprawdzˊmy, ile hkm​ ma 1 sm​:

a=5⋅2−2+4⋅3−2−6⋅2−3   =5⋅221​+4⋅321​−6⋅231​=5⋅41​+4⋅91​−6⋅81​=