II gimnazjum

Matematyka z plusem 2

Rozwiązanie 1

a) 311⋅2427⋅315​=27−4⋅315−11=23⋅34   311⋅2427⋅315​&lt;23⋅35

a) 4⋅(3423​)10=22⋅(34)10(23)10​   =22⋅34⋅1023⋅10​=   22⋅340230​=   34022+30​=340232​

8100⋅41​=(23)100⋅(21​)2=

a) a=1210=122⋅5=(122)5=1445&gt;245

a) 27⋅37:65=(2⋅3)7:65=67:65=67−5=62=36 b) (−21​)3⋅(−21​)4⋅27=(−21​)3+4⋅27=(−21​)7⋅27=(−21​⋅2)7=(−1)7=−1

a) 10455​=5⋅10454​=5⋅(105​)4=5⋅(21​)4=5⋅161​=165​   b) 223444​=44⋅223443​=44⋅(2244​)3=44⋅23=44⋅8=352

pod koniec 1859: 22 pod koniec 1860: 22*2  pod koniec 1861: 22*2*2=22*22   pod koniec 1862: 22*23

a) 41​=(21​)2=2−2 a=−2   b) 27=33=(31​)−3

a) (71​)−1=7   b) (71​)−11​=71​   c) <svg xmlns="http://www.w3.org/2000/svg" width='0.875em' height='3.600em' viewBox='0 0 875 3600'><path d='M863,9c0,-2,-2,-5,-6,-9c0,0,-17,0,-17,0c-12.7,0,-19.3,0.3,-20,1
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