6 szkoły podstawowej

Matematyka z plusem 6

Rozwiązanie 1

a) Liczby większe od -0,6 to:  −21​,   −74​

Aby określić odległość na osi liczbowej między liczbami a i b należy od większej z tych liczb odjąć mniejszą.  a)  −7,5 &lt; 8 Mamy więc:

a)  6−14−(−5)−2=−8+5−2=−3−2=−5

a)  373​−7=−(7−373​)=−(677​−373​)=−374​

a)  6,75+<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMinYMin slice'><path d='M0 6l6-6h17c12.688 0 19.313.3 20 1 4 4 7.313 8.3 10 13
 35.313 51.3 80.813 93.8 136.5 127.5 55.688 33.7 117.188 55.8 184.5 66.5.688
 0 2 .3 4 1 18.688 2.7 76 4.3 172 5h399450v120H429l-6-1c-124.688-8-235-61.7
-331-161C60.687 138.7 32.312 99.3 7 54L0 41V6z'/></svg><svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMidYMin slice'><path d='M199572 214
c100.7 8.3 195.3 44 280 108 55.3 42 101.7 93 139 153l9 14c2.7-4 5.7-8.7 9-14
 53.3-86.7 123.7-153 211-199 66.7-36 137.3-56.3 212-62h199568v120H200432c-178.3
 11.7-311.7 78.3-403 201-6 8-9.7 12-11 12-.7.7-6.7 1-18 1s-17.3-.3-18-1c-1.3 0
-5-4-11-12-44.7-59.3-101.3-106.3-170-141s-145.3-54.3-229-60H0V214z'/></svg><svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMaxYMin slice'><path d='M399994 0l6 6v35l-6 11c-56 104-135.3 181.3-238 232-57.3
 28.7-117 45-179 50H-300V214h399897c43.3-7 81-15 113-26 100.7-33 179.7-91 237
-174 2.7-5 6-9 10-13 .7-1 7.3-1 20-1h17z'/></svg>-(-​​(-(-(-5,197))))=6,75−<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMinYMin slice'><path d='M0 6l6-6h17c12.688 0 19.313.3 20 1 4 4 7.313 8.3 10 13
 35.313 51.3 80.813 93.8 136.5 127.5 55.688 33.7 117.188 55.8 184.5 66.5.688
 0 2 .3 4 1 18.688 2.7 76 4.3 172 5h399450v120H429l-6-1c-124.688-8-235-61.7
-331-161C60.687 138.7 32.312 99.3 7 54L0 41V6z'/></svg><svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMidYMin slice'><path d='M199572 214
c100.7 8.3 195.3 44 280 108 55.3 42 101.7 93 139 153l9 14c2.7-4 5.7-8.7 9-14
 53.3-86.7 123.7-153 211-199 66.7-36 137.3-56.3 212-62h199568v120H200432c-178.3
 11.7-311.7 78.3-403 201-6 8-9.7 12-11 12-.7.7-6.7 1-18 1s-17.3-.3-18-1c-1.3 0
-5-4-11-12-44.7-59.3-101.3-106.3-170-141s-145.3-54.3-229-60H0V214z'/></svg><svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMaxYMin slice'><path d='M399994 0l6 6v35l-6 11c-56 104-135.3 181.3-238 232-57.3
 28.7-117 45-179 50H-300V214h399897c43.3-7 81-15 113-26 100.7-33 179.7-91 237
-174 2.7-5 6-9 10-13 .7-1 7.3-1 20-1h17z'/></svg>+(-​​(-(-5,197)))=6,75+<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMinYMin slice'><path d='M0 6l6-6h17c12.688 0 19.313.3 20 1 4 4 7.313 8.3 10 13
 35.313 51.3 80.813 93.8 136.5 127.5 55.688 33.7 117.188 55.8 184.5 66.5.688
 0 2 .3 4 1 18.688 2.7 76 4.3 172 5h399450v120H429l-6-1c-124.688-8-235-61.7
-331-161C60.687 138.7 32.312 99.3 7 54L0 41V6z'/></svg><svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMidYMin slice'><path d='M199572 214
c100.7 8.3 195.3 44 280 108 55.3 42 101.7 93 139 153l9 14c2.7-4 5.7-8.7 9-14
 53.3-86.7 123.7-153 211-199 66.7-36 137.3-56.3 212-62h199568v120H200432c-178.3
 11.7-311.7 78.3-403 201-6 8-9.7 12-11 12-.7.7-6.7 1-18 1s-17.3-.3-18-1c-1.3 0
-5-4-11-12-44.7-59.3-101.3-106.3-170-141s-145.3-54.3-229-60H0V214z'/></svg><svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMaxYMin slice'><path d='M399994 0l6 6v35l-6 11c-56 104-135.3 181.3-238 232-57.3
 28.7-117 45-179 50H-300V214h399897c43.3-7 81-15 113-26 100.7-33 179.7-91 237
-174 2.7-5 6-9 10-13 .7-1 7.3-1 20-1h17z'/></svg>-(-​​(-5,197))=

a)  343​−(7−821​)−6=343​−(−121​)−6=343​+121​−6=343​+142​−6=445​−6=541​−6=−43​

Obliczamy ile wynosi wartość każdego z podanych wyrażeń.  Następnie sprawdzamy, czy jest ona większa od -1 i mniejsza od -0,5.  a)  (21​−141​)−0,3=(42​−141​)−0,3=−(141​−42​)−0,3=−(45​−42​)−0,3=−43​−0,3=−0,75−0,3=−1,05

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