Odpowiedź :
- Działanie A)
[tex](\frac{3}{7}+\frac{2}{5}):4\frac{1}{7}=(\frac{3\cdot5}{7\cdot5}+\frac{2\cdot7}{5\cdot7}):\frac{4\cdot7+1}{7}=(\frac{15}{35}+\frac{14}{35}):\frac{28+1}{7}=\\\\=\frac{29}{35}:\frac{29}{7}=\frac{\not29}{\not35_5}\cdot\frac{\not7^1}{\not29}=\frac{1}{5}[/tex]
- Działanie B)
[tex]3\frac{3}{4}\cdot(\frac{2}{5}-\frac{1}{2})=\frac{3\cdot4+3}{4}\cdot(\frac{2\cdot2}{5\cdot2}-\frac{1\cdot5}{2\cdot5})=\frac{12+3}{4}\cdot(\frac{4}{10}-\frac{5}{10})=\frac{\not15^3}{4}\cdot(-\frac{1}{\not10_2})=-\frac{3}{8}\\[/tex]
[tex]A) \ (\dfrac{3}{7}+\dfrac{2}{5}):4\dfrac{1}{7} =(\dfrac{15}{35}+\dfrac{14}{35}):\dfrac{29}{7} = \dfrac{29}{35}\cdot\dfrac{7}{29} = \dfrac{7}{35} = \dfrac{1}{5}[/tex]
[tex]B) \ 3\dfrac{3}{4}\dcdot(\dfrac{2}{5}-\dfrac{1}{2}) = \dfrac{15}{4}\dcdot(\dfrac{4}{10}-\dfrac{5}{10}) = \dfrac{15}{4}\cdot(-\frac{1}{10}) = \dfrac{3}{4}\cdot(-\dfrac{1}{2}) = -\dfrac{3}{8}[/tex]