Cześć ;-)
Od a) do c)
[tex]a] \ \frac{3}{8}+\frac{2}{5}=\frac{15}{40}+\frac{16}{40}=\frac{31}{40}\\\\b] \ 1\frac{2}{7}+3\frac{4}{9}=1\frac{18}{63}+3\frac{28}{63}=4\frac{36}{63}=4\frac{4}{7}\\\\c] \ \frac{5}{6}+\frac{7}{8}=\frac{20}{24}+\frac{21}{24}=\frac{41}{24}=1\frac{17}{24}[/tex]
Od d) do f)
[tex]d] \ -3\frac{2}{7}+7\frac{6}{7}=7\frac{6}{7}-3\frac{2}{7}=4\frac{4}{7}\\\\e] \ \frac{1}{6}-\frac{2}{18}=\frac{3}{18}-\frac{2}{18}=\frac{1}{18}\\\\f] \ \frac{3}{\not4_2}\cdot(-\frac{\not2^1}{7})=-\frac{3}{14}[/tex]
Od g) do i)
[tex]g] \ \frac{5}{13}\cdot2\frac{9}{15}=\frac{\not5^1}{\not13_1}\cdot\frac{\not39^3}{\not15_3}=\frac{3}{3}=1\\\\h] \ \frac{8}{15}:\frac{4}{5}=\frac{\not8^2}{\not15_3}\cdot\frac{\not5^1}{\not4_1}=\frac{2}{3}\\\\i] \ -\frac{16}{27}:\frac{2}{3}=-\frac{\not16^8}{\not27_9}\cdot\frac{\not3^1}{\not2_1}=-\frac{8}{9}[/tex]
Pozdrawiam!
