Skracanie ułamków – Zadanie 3

Doprowadź wyrażenia do najprostszej postaci

a) \(\dfrac{4}{24}\cdot \dfrac{12}{14}\)

b) \(\dfrac{2}{26}\cdot \dfrac{16}{10}\)

c) \(\dfrac{22}{30}\cdot \dfrac{5}{11}\)

d) \(\dfrac{8}{28}\cdot \dfrac{7}{12}\)

e) \(\dfrac{19}{20}\cdot \dfrac{15}{38}\)

Rozwiązanie
a)
\(\dfrac{4}{24}\cdot \dfrac{12}{14}=\dfrac{4\cdot 12}{24\cdot 14}=\dfrac{4\cdot 1}{2\cdot 14}=\dfrac{1\cdot 1}{1\cdot 7}=\dfrac{1}{7}\)

b)
\(\dfrac{2}{26}\cdot \dfrac{16}{10}=\dfrac{2\cdot 16}{26\cdot 10}=\dfrac{1\cdot 8}{13\cdot 5}=\dfrac{8}{65}\)

c)
\(\dfrac{22}{30}\cdot \dfrac{5}{11}=\dfrac{22\cdot 5}{30\cdot 11}=\dfrac{2\cdot 1}{6\cdot 1}=\dfrac{2}{6}==\dfrac{1}{3}\)

d)
\(\dfrac{8}{28}\cdot \dfrac{7}{12}=\dfrac{8\cdot 7}{28\cdot 12}=\dfrac{2\cdot 1}{4\cdot 3}=\dfrac{1\cdot 1}{2\cdot 3}=\dfrac{1}{6}\)

e)
\(\dfrac{19}{20}\cdot \dfrac{15}{38}=\dfrac{19\cdot 15}{20\cdot 38}=\dfrac{1\cdot 3}{4\cdot 2}=\dfrac{3}{8}\)