Temat: Działania na ułamkach
Rozwiązania poniżej ;-)
a)
[tex]12-4\dfrac{1}{8}=11\dfrac{8}{8}-4\dfrac{1}{8}=\boxed{7\frac{7}{8}}[/tex]
b)
[tex]\dfrac{11}{30}-\dfrac{5}{6}=\dfrac{11}{30}-\dfrac{5\cdot5}{6\cdot5}=\dfrac{11}{30}-\dfrac{25}{30}=-\dfrac{14}{30}=\boxed{-\frac{7}{15}}\\[/tex]
c)
[tex]5\dfrac{1}{14}-2\dfrac{1}{4}=5\dfrac{1\cdot2}{14\cdot2}-2\dfrac{1\cdot7}{4\cdot7}=15\dfrac{2}{28}-2\dfrac{7}{28}=14\dfrac{30}{28}-2\dfrac{7}{28}=\boxed{12\frac{23}{28}}\\[/tex]
d)
[tex]12\dfrac{2}{3}-8\dfrac{3}{4}=12\dfrac{2\cdot4}{3\cdot4}-8\dfrac{3\cdot3}{4\cdot3}=12\dfrac{8}{12}-8\dfrac{9}{12}=11\dfrac{20}{12}-8\dfrac{9}{12}=\boxed{3\frac{11}{12}}\\[/tex]
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Temat: Działania na ułamkach
Rozwiązania poniżej ;-)
a)
[tex]12-4\dfrac{1}{8}=11\dfrac{8}{8}-4\dfrac{1}{8}=\boxed{7\frac{7}{8}}[/tex]
b)
[tex]\dfrac{11}{30}-\dfrac{5}{6}=\dfrac{11}{30}-\dfrac{5\cdot5}{6\cdot5}=\dfrac{11}{30}-\dfrac{25}{30}=-\dfrac{14}{30}=\boxed{-\frac{7}{15}}\\[/tex]
c)
[tex]5\dfrac{1}{14}-2\dfrac{1}{4}=5\dfrac{1\cdot2}{14\cdot2}-2\dfrac{1\cdot7}{4\cdot7}=15\dfrac{2}{28}-2\dfrac{7}{28}=14\dfrac{30}{28}-2\dfrac{7}{28}=\boxed{12\frac{23}{28}}\\[/tex]
d)
[tex]12\dfrac{2}{3}-8\dfrac{3}{4}=12\dfrac{2\cdot4}{3\cdot4}-8\dfrac{3\cdot3}{4\cdot3}=12\dfrac{8}{12}-8\dfrac{9}{12}=11\dfrac{20}{12}-8\dfrac{9}{12}=\boxed{3\frac{11}{12}}\\[/tex]