Hej!
Od a) do c) :
[tex]a] \ (1\frac{1}{7}+\frac{2}{5})\cdot3=(1\frac{5}{35}+\frac{14}{35})\cdot2=1\frac{19}{35}\cdot2=\frac{54}{35}\cdot2=\frac{108}{35}=3\frac{3}{35}\\\\b] \ 2\frac{1}{4}:\frac{3}{5}=\frac{\not9^3}{4}\cdot\frac{5}{\not3_1}=\frac{15}{4}=3\frac{3}{4}\\\\c] \ 3\frac{1}{5}\cdot7=\frac{16}{5}\cdot7=\frac{112}{5}=22\frac{2}{5}[/tex]
Od d) do f) :
[tex]d] \ 8\frac{1}{2}:34=\frac{\not17^1}{2}\cdot\frac{1}{\not34_2}=\frac{1}{4}\\\\e] \ 10\frac{1}{2}-2\frac{3}{7}\cdot2=10\frac{1}{2}-4\frac{6}{7}=10\frac{7}{14}-4\frac{12}{14}=9\frac{21}{14}-4\frac{12}{14}=5\frac{9}{14}\\\\f] \ (-\frac{\not3^1}{\not5_1})\cdot(-\frac{\not20^4}{\not27_3})\cdot\frac{1}{9}=\frac{4}{3}\cdot\frac{1}{9}=\frac{4}{27}[/tex]
