I gimnazjum

Matematyka wokół nas 1

[−2,25−(−487​)]⋅8=(−241​+487​)⋅8= [−2,25−(−487​)]⋅8=(−241​+487​)⋅8= (−282​+487​)⋅8=285​⋅8=821​⋅8=21

Rozwiązanie 1

4,2⋅331​⋅(−6)+[165​−2,75​]=

3383​<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='2.48em' viewBox='0 0 400000 2592' preserveAspectRatio='xMinYMin slice'><path d='M424,2478
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s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185
c-2,6,-10,9,-24,9
c-8,0,-12,-0.7,-12,-2z M1001 80
h400000v40h-400000z'/></svg>​+(−3)2⋅x−(1−241​)2=13167​ 3827​<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='2.48em' viewBox='0 0 400000 2592' preserveAspectRatio='xMinYMin slice'><path d='M424,2478
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s209,968,209,968c0,-2,84.7,-361.7,254,-1079c169.3,-717.3,254.7,-1077.7,256,-1081
l0 -0c4,-6.7,10,-10,18,-10 H400000
v40H1014.6
s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185
c-2,6,-10,9,-24,9
c-8,0,-12,-0.7,-12,-2z M1001 80
h400000v40h-400000z'/></svg>​+9⋅x−(−141​)2=13167​

3,2:1151​⋅(−65​)=3102​:1516​⋅(−65​)=1032​⋅1615​⋅(−65​)=21​⋅11​⋅(−15​)=−25​

1169​<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='2.48em' viewBox='0 0 400000 2592' preserveAspectRatio='xMinYMin slice'><path d='M424,2478
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h400000v40h-400000z'/></svg>​⋅38<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>​=

1058−134 jest: 6 ponieważ ostatnia cyfrą liczby 1058 jest 0 i przy odejmowaniu pisemnym będzie 10-4=6