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[tex]1\frac{2}{7}\cdot2\frac{4}{5}-6\frac{2}{5}:3\frac{1}{2}=\frac{1\cdot7+2}{7}\cdot\frac{2\cdot5+4}{5}-\frac{6\cdot5+2}{5}:\frac{3\cdot2+1}{2}=\\\\=\frac{7+2}{7}\cdot\frac{10+4}{5}-\frac{30+2}{5}:\frac{6+1}{2}=\frac{9}{\not7_1}\cdot\frac{\not14^2}{5}-\frac{32}{5}:\frac{7}{2}=\\\\=\frac{18}{5}-\frac{32}{5}\cdot\frac{2}{7}=\frac{18}{5}-\frac{64}{35}=\frac{18\cdot7}{5\cdot7}-\frac{64}{35}=\frac{126}{35}-\frac{64}{35}=\frac{62}{35}=1\frac{27}{35}[/tex]
Odpowiedź:
1 i 2/7 * 2 i 4/5 - 6 i 2/5 : 3 i 1/2 = 9/7 * 14/5 - 32/5 : 7/2 = 18/5 - 32/5 * 2/7 = 18/5 - 64/35 = 126/35 - 64/35 = 62/35 = 1 i 27/35