[tex]\textbf{12.}\\\\\textbf{a)}\frac{5}{7}: 3=\frac{5}{21}\\\\6: 2 \frac{1}{5}=2 \frac{8}{11} \\\\7 \frac{1}{6}: 9=\frac{43}{54}[/tex] [tex]\textbf{b)}\ \frac{2}{3}: \frac{3}{5}=1 \frac{1}{9}\\\\3 \frac{3}{7}: \frac{5}{8}=\frac{192}{35}\\\\1 \frac{3}{4}: 1 \frac{3}{5}=\frac{35}{32}[/tex]
[tex]\textbf{13.}\\\\\textbf{a) }\ \dfrac{\frac{2}{5}}{\frac{4}{15}}=1\frac{1}{2}\\\\\\\textbf{b)}\ \dfrac{3}{\frac{1}{5}}=15\\\\\\\textbf{c)}\ \dfrac{\frac{3}{4}}{5}=\frac{3}{20}[/tex]
[tex]\textbf{14.}\ \\\\\textbf{a)}\ \frac{2}{7} \ \ z\ \ 14\ kg\ \ to\ \ 4\ kg\\\\\textbf{b)}\ \frac{3}{5} \ \ z\ \ 60\ zl\ \ to\ \ 36 \ zl\\\\\textbf{c)}\ \frac{3}{5} \ \ z\ \ 30\ \ l \ to\ \ 18\ l[/tex]
[tex]\textbf{15.}\ \\\\\textbf{a)}\ \frac{3}{7} \ \ liczby\ \ to \ \ 12 \\\\\textbf{b)}\ \frac{1}{3} \ \ liczby \ \ 3\frac{3}{4}\ \ to\ \ 1\frac{1}{4}\\\\\textbf{c)}\ 1\frac{1}{7} \ \ liczby\ \ 2\frac{1}{4}\ \ to\ \2\frac{4}{7}[/tex]
[tex]\textbf{16.}\\\\\textbf{a)}\ 1 \frac{3}{4} \cdot 1 \frac{1}{7}-3: 4 \frac{1}{2}=1 \frac{1}{3}\\\\\textbf{b)}\ 1 \frac{5}{7} \cdot\left(2 \frac{3}{4}-1 \frac{2}{3}: 4\right)=4[/tex]
[tex]\textbf{c)}\ 3 \frac{1}{3} \cdot\left(2 \frac{1}{4}-1 \frac{1}{2}: 5\right)=\frac{13}{20}\\\\[/tex]
[tex]\textbf{d)}\ \dfrac{3 \frac{1}{4}+\frac{1}{2}}{3 \frac{1}{3} \cdot\left(1 \frac{5}{6}-1 \frac{1}{12}\right)}=1 \frac{1}{2}$[/tex]
W tym zadaniu musimy obliczyć wyrażenia, w których pojawiają się ułamki. Przypomnijmy kilka zasad:
Pamiętajmy też o kolejności wykonywania działań:
nawiasy ⇒ potęgowanie/pierwiastkowanie ⇒ mnożenie/dzielenie ⇒ dodawanie/odejmowanie
Ważne jest również, by zwracać uwagę na znaki:
Pamiętajmy, że jeśli chcemy obliczyć ułamek z danej liczby, to musimy wykonać mnożenie.
Szczegółowe rozwiązanie
[tex]\textbf{12.}\\\\\textbf{a)}\frac{5}{7}: 3=\frac{5}{7} \cdot \frac{1}{3}=\frac{5}{21}\\\\6: 2 \frac{1}{5}=6: \frac{11}{5}=\frac{6}{1} \cdot \frac{5}{11}=\frac{30}{11}=2 \frac{8}{11} \\\\7 \frac{1}{6}: 9=\frac{43}{6} \cdot \frac{1}{9}=\frac{43}{54}\end{aligned}$$[/tex] [tex]\textbf{b)}\ \frac{2}{3}: \frac{3}{5}=\frac{2}{3} \cdot \frac{5}{3}=\frac{10}{9}=1 \frac{1}{9}\\\\3 \frac{3}{7}: \frac{5}{8}=\frac{24}{7} \cdot \frac{8}{5}=\frac{192}{35}\\\\1 \frac{3}{4}: 1 \frac{3}{5}=\frac{7}{4}: \frac{8}{5}=\frac{7}{4} \cdot \frac{5}{8}=\frac{35}{32}[/tex]
[tex]\textbf{13.}\\\\\textbf{a) }\ \dfrac{\frac{2}{5}}{\frac{4}{15}}=\frac{\not2}{\not5} \cdot \frac{\not15}{\not4}=\frac{1 \cdot 3}{1 \cdot 2}=\frac{3}{2}=1\frac{1}{2}\\\\\\\textbf{b)}\ \dfrac{3}{\frac{1}{5}}=3 \cdot 5=15\\\\\\\textbf{c)}\ \dfrac{\frac{3}{4}}{5}=\frac{3}{4} \cdot \frac{1}{5}=\frac{3}{20}[/tex]
[tex]\textbf{14.}\ \\\\\textbf{a)}\ \frac{2}{7} \cdot 14=\frac{2}{1}\cdot2=4\ [kg]\\\\\textbf{b)}\ \frac{3}{5} \cdot 60=\frac{3}{1} \cdot 12=36 \ [zl]\\\\\textbf{c)}\ \frac{3}{5} \cdot 30=\frac{3}{1}\cdot6=18\ [l][/tex]
[tex]\textbf{15.}\ \\\\\textbf{a)}\ \frac{3}{7} \cdot 28=\frac{3}{1}\cdot4=12\\\\\textbf{b)}\ \frac{1}{3} \cdot 3\frac{3}{4}=\frac{1}{3}\cdot \frac{15}{4}=\frac{1}{1}\cdot\frac{5}{4}=\frac{5}{4}=1\frac{1}{4}\\\\\textbf{c)}\ 1\frac{1}{7} \cdot 2\frac{1}{4}=\frac{8}{7}\cdot\frac{9}{4}=\frac{2}{7}\cdot\frac{9}{1}=\frac{18}{7}=2\frac{4}{7}[/tex]
[tex]\textbf{16.}\\\\\textbf{a)}\ 1 \frac{3}{4} \cdot 1 \frac{1}{7}-3: 4 \frac{1}{2}=\frac{7}{4} \cdot \frac{8}{7}-3: \frac{9}{2}=\frac{2}{1}-3 \cdot \frac{2}{9}=2-\frac{2}{3}=1 \frac{1}{3}\\\\\textbf{b)}\ 1 \frac{5}{7} \cdot\left(2 \frac{3}{4}-1 \frac{2}{3}: 4\right)=\frac{12}{7} \cdot\left(\frac{11}{4}-\frac{5}{3} \cdot \frac{1}{4}\right)=\frac{12}{7} \cdot\left(\frac{11}{4}-\frac{5}{12}\right)=\frac{12}{7} \cdot\left(\frac{33}{12}-\frac{5}{12}\right)=\\\\=\frac{12}{7} \cdot \frac{28}{12}=\frac{4}{1}=4[/tex]
[tex]\textbf{c)}\ 3 \frac{1}{3} \cdot\left(2 \frac{1}{4}-1 \frac{1}{2}: 5\right)=\frac{10}{3} \cdot\left(\frac{9}{4}-\frac{3}{2} \cdot \frac{1}{5}\right)=\frac{10}{3} \cdot\left(\frac{9}{4}-\frac{3}{10}\right)=\frac{10}{3} \cdot\left(\frac{45}{20}-\frac{6}{20}\right)=\\\\=\frac{10}{30} \cdot \frac{39}{20}=\frac{1 \cdot 13}{10 \cdot 2}=\frac{13}{20}\\\\[/tex]
[tex]\textbf{d)}\ \dfrac{3 \frac{1}{4}+\frac{1}{2}}{3 \frac{1}{3} \cdot\left(1 \frac{5}{6}-1 \frac{1}{12}\right)}=\dfrac{\frac{13}{4}+\frac{2}{4}}{\frac{10}{3} \cdot\left(\frac{11}{6}-\frac{13}{12}\right)}=\dfrac{\frac{15}{4}}{\frac{10}{3} \cdot\left(\frac{22}{12}-\frac{13}{12}\right)}=\dfrac{\frac{15}{4}}{\frac{10}{3} \cdot \frac{9}{12}}=\dfrac{\frac{15}{4}}{\frac{15}{6}}=\\\\=\frac{15}{4} \cdot \frac{6}{15}=\frac{1 \cdot 3}{2 \cdot 1}=\frac{3}{2}=1 \frac{1}{2}$[/tex]
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[tex]\textbf{12.}\\\\\textbf{a)}\frac{5}{7}: 3=\frac{5}{21}\\\\6: 2 \frac{1}{5}=2 \frac{8}{11} \\\\7 \frac{1}{6}: 9=\frac{43}{54}[/tex] [tex]\textbf{b)}\ \frac{2}{3}: \frac{3}{5}=1 \frac{1}{9}\\\\3 \frac{3}{7}: \frac{5}{8}=\frac{192}{35}\\\\1 \frac{3}{4}: 1 \frac{3}{5}=\frac{35}{32}[/tex]
[tex]\textbf{13.}\\\\\textbf{a) }\ \dfrac{\frac{2}{5}}{\frac{4}{15}}=1\frac{1}{2}\\\\\\\textbf{b)}\ \dfrac{3}{\frac{1}{5}}=15\\\\\\\textbf{c)}\ \dfrac{\frac{3}{4}}{5}=\frac{3}{20}[/tex]
[tex]\textbf{14.}\ \\\\\textbf{a)}\ \frac{2}{7} \ \ z\ \ 14\ kg\ \ to\ \ 4\ kg\\\\\textbf{b)}\ \frac{3}{5} \ \ z\ \ 60\ zl\ \ to\ \ 36 \ zl\\\\\textbf{c)}\ \frac{3}{5} \ \ z\ \ 30\ \ l \ to\ \ 18\ l[/tex]
[tex]\textbf{15.}\ \\\\\textbf{a)}\ \frac{3}{7} \ \ liczby\ \ to \ \ 12 \\\\\textbf{b)}\ \frac{1}{3} \ \ liczby \ \ 3\frac{3}{4}\ \ to\ \ 1\frac{1}{4}\\\\\textbf{c)}\ 1\frac{1}{7} \ \ liczby\ \ 2\frac{1}{4}\ \ to\ \2\frac{4}{7}[/tex]
[tex]\textbf{16.}\\\\\textbf{a)}\ 1 \frac{3}{4} \cdot 1 \frac{1}{7}-3: 4 \frac{1}{2}=1 \frac{1}{3}\\\\\textbf{b)}\ 1 \frac{5}{7} \cdot\left(2 \frac{3}{4}-1 \frac{2}{3}: 4\right)=4[/tex]
[tex]\textbf{c)}\ 3 \frac{1}{3} \cdot\left(2 \frac{1}{4}-1 \frac{1}{2}: 5\right)=\frac{13}{20}\\\\[/tex]
[tex]\textbf{d)}\ \dfrac{3 \frac{1}{4}+\frac{1}{2}}{3 \frac{1}{3} \cdot\left(1 \frac{5}{6}-1 \frac{1}{12}\right)}=1 \frac{1}{2}$[/tex]
Działania na ułamkach
W tym zadaniu musimy obliczyć wyrażenia, w których pojawiają się ułamki. Przypomnijmy kilka zasad:
Pamiętajmy też o kolejności wykonywania działań:
nawiasy ⇒ potęgowanie/pierwiastkowanie ⇒ mnożenie/dzielenie ⇒ dodawanie/odejmowanie
Ważne jest również, by zwracać uwagę na znaki:
Pamiętajmy, że jeśli chcemy obliczyć ułamek z danej liczby, to musimy wykonać mnożenie.
Szczegółowe rozwiązanie
[tex]\textbf{12.}\\\\\textbf{a)}\frac{5}{7}: 3=\frac{5}{7} \cdot \frac{1}{3}=\frac{5}{21}\\\\6: 2 \frac{1}{5}=6: \frac{11}{5}=\frac{6}{1} \cdot \frac{5}{11}=\frac{30}{11}=2 \frac{8}{11} \\\\7 \frac{1}{6}: 9=\frac{43}{6} \cdot \frac{1}{9}=\frac{43}{54}\end{aligned}$$[/tex] [tex]\textbf{b)}\ \frac{2}{3}: \frac{3}{5}=\frac{2}{3} \cdot \frac{5}{3}=\frac{10}{9}=1 \frac{1}{9}\\\\3 \frac{3}{7}: \frac{5}{8}=\frac{24}{7} \cdot \frac{8}{5}=\frac{192}{35}\\\\1 \frac{3}{4}: 1 \frac{3}{5}=\frac{7}{4}: \frac{8}{5}=\frac{7}{4} \cdot \frac{5}{8}=\frac{35}{32}[/tex]
[tex]\textbf{13.}\\\\\textbf{a) }\ \dfrac{\frac{2}{5}}{\frac{4}{15}}=\frac{\not2}{\not5} \cdot \frac{\not15}{\not4}=\frac{1 \cdot 3}{1 \cdot 2}=\frac{3}{2}=1\frac{1}{2}\\\\\\\textbf{b)}\ \dfrac{3}{\frac{1}{5}}=3 \cdot 5=15\\\\\\\textbf{c)}\ \dfrac{\frac{3}{4}}{5}=\frac{3}{4} \cdot \frac{1}{5}=\frac{3}{20}[/tex]
[tex]\textbf{14.}\ \\\\\textbf{a)}\ \frac{2}{7} \cdot 14=\frac{2}{1}\cdot2=4\ [kg]\\\\\textbf{b)}\ \frac{3}{5} \cdot 60=\frac{3}{1} \cdot 12=36 \ [zl]\\\\\textbf{c)}\ \frac{3}{5} \cdot 30=\frac{3}{1}\cdot6=18\ [l][/tex]
[tex]\textbf{15.}\ \\\\\textbf{a)}\ \frac{3}{7} \cdot 28=\frac{3}{1}\cdot4=12\\\\\textbf{b)}\ \frac{1}{3} \cdot 3\frac{3}{4}=\frac{1}{3}\cdot \frac{15}{4}=\frac{1}{1}\cdot\frac{5}{4}=\frac{5}{4}=1\frac{1}{4}\\\\\textbf{c)}\ 1\frac{1}{7} \cdot 2\frac{1}{4}=\frac{8}{7}\cdot\frac{9}{4}=\frac{2}{7}\cdot\frac{9}{1}=\frac{18}{7}=2\frac{4}{7}[/tex]
[tex]\textbf{16.}\\\\\textbf{a)}\ 1 \frac{3}{4} \cdot 1 \frac{1}{7}-3: 4 \frac{1}{2}=\frac{7}{4} \cdot \frac{8}{7}-3: \frac{9}{2}=\frac{2}{1}-3 \cdot \frac{2}{9}=2-\frac{2}{3}=1 \frac{1}{3}\\\\\textbf{b)}\ 1 \frac{5}{7} \cdot\left(2 \frac{3}{4}-1 \frac{2}{3}: 4\right)=\frac{12}{7} \cdot\left(\frac{11}{4}-\frac{5}{3} \cdot \frac{1}{4}\right)=\frac{12}{7} \cdot\left(\frac{11}{4}-\frac{5}{12}\right)=\frac{12}{7} \cdot\left(\frac{33}{12}-\frac{5}{12}\right)=\\\\=\frac{12}{7} \cdot \frac{28}{12}=\frac{4}{1}=4[/tex]
[tex]\textbf{c)}\ 3 \frac{1}{3} \cdot\left(2 \frac{1}{4}-1 \frac{1}{2}: 5\right)=\frac{10}{3} \cdot\left(\frac{9}{4}-\frac{3}{2} \cdot \frac{1}{5}\right)=\frac{10}{3} \cdot\left(\frac{9}{4}-\frac{3}{10}\right)=\frac{10}{3} \cdot\left(\frac{45}{20}-\frac{6}{20}\right)=\\\\=\frac{10}{30} \cdot \frac{39}{20}=\frac{1 \cdot 13}{10 \cdot 2}=\frac{13}{20}\\\\[/tex]
[tex]\textbf{d)}\ \dfrac{3 \frac{1}{4}+\frac{1}{2}}{3 \frac{1}{3} \cdot\left(1 \frac{5}{6}-1 \frac{1}{12}\right)}=\dfrac{\frac{13}{4}+\frac{2}{4}}{\frac{10}{3} \cdot\left(\frac{11}{6}-\frac{13}{12}\right)}=\dfrac{\frac{15}{4}}{\frac{10}{3} \cdot\left(\frac{22}{12}-\frac{13}{12}\right)}=\dfrac{\frac{15}{4}}{\frac{10}{3} \cdot \frac{9}{12}}=\dfrac{\frac{15}{4}}{\frac{15}{6}}=\\\\=\frac{15}{4} \cdot \frac{6}{15}=\frac{1 \cdot 3}{2 \cdot 1}=\frac{3}{2}=1 \frac{1}{2}$[/tex]
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