Kiedy mnożymy dwa ułamki, to licznik mnożymy przez licznik, a mianownik przez mianownik. $\frac{a}{b}\cdot \frac{c}{d}=\frac{a\cdot c}{b\cdot d}$

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Wykonajmy obliczenia.

a)

$\frac{{\cancel{3}}^1}{7}\cdot \frac{5}{{\cancel{9}}_3}=\frac{1\cdot 5}{7\cdot 3}=\frac{5}{21}$

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$\frac{{\cancel{4}}^1}{3}\cdot \frac{7}{{\cancel{16}}_4}=\frac{1\cdot 7}{3\cdot 4}=\frac{7}{12}$

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$\frac{{\cancel{6}}^2}{11}\cdot \frac{5}{{\cancel{9}}_3}=\frac{2\cdot 5}{11\cdot 3}=\frac{10}{33}$

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b)

$\frac{{\cancel{8}}^2}{{\cancel{21}}_3}\cdot \frac{{\cancel{7}}^1}{{\cancel{12}}_3}=\frac{2\cdot 1}{3\cdot 3}=\frac{2}{9}$

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$\frac{{\cancel{5}}^1}{{\cancel{6}}_2}\cdot \frac{{\cancel{9}}^3}{{\cancel{20}}_4}=\frac{1\cdot 3}{2\cdot 4}=\frac{3}{8}$

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$\frac{{\cancel{4}}^1}{{\cancel{7}}_1}\cdot \frac{{\cancel{21}}^3}{{\cancel{36}}_9}=\frac{1\cdot 3}{1\cdot 9}=\frac{3}{9}=\frac{1}{3}$

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c)

$1\frac{1}{3}\cdot 2\frac{1}{4}=\frac{{\cancel{4}}^1}{{\cancel{3}}_1}\cdot \frac{{\cancel{9}}^3}{{\cancel{4}}_1}=\frac{3}{1}=3$

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$1\frac{1}{2}\cdot 1\frac{5}{9}=\frac{{\cancel{3}}^1}{{\cancel{2}}_1}\cdot \frac{{\cancel{14}}^7}{{\cancel{9}}_3}=\frac{7}{3}=2\frac{1}{3}$

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$2\frac{5}{8}\cdot 1\frac{1}{3}=\frac{{\cancel{21}}^7}{{\cancel{8}}_2}\cdot \frac{{\cancel{4}}^1}{{\cancel{3}}_1}=\frac{7\cdot 1}{2\cdot 1}=\frac{7}{2}=3\frac{1}{2}$