Cześć ;-)
Zadanie 14
Od a) do d)
[tex]1,16+\frac{1}{4}=1,16+\frac{25}{100}=1,16+0,25=1,41\\\\3,9+\frac{1}{25}=3,9+\frac{4}{100}=3,9+0,04=3,94\\\\65,8-\frac{3}{20}=65,8-\frac{15}{100}=65,8-0,15=65,65\\\\3,65-1\frac{2}{5}=3,65-1\frac{4}{10}=3,65-1,4=2,25[/tex]
Od e) do h)
[tex]\frac{7}{8}-0,15=\frac{875}{1000}-0,15=0,875-0,15=0,725\\\\17\frac{3}{4}+1,7=17\frac{75}{100}+1,7=17,75+1,7=19,45\\\\2,5-\frac{2}{5}=2,5-\frac{4}{10}=2,5-0,4=2,1\\\\3\frac{1}{8}-0,2=3\frac{125}{100}-0,2=3,125-0,2=2,925[/tex]
Zadanie 15
Od a) do c)
[tex]3,6+\frac{2}{3}=3\frac{6}{10}+\frac{2}{3}=3\frac{3}{5}+\frac{2}{3}=3\frac{9}{15}+\frac{10}{15}=3\frac{19}{15}=4\frac{4}{15}\\\\4\frac{1}{8}-1,11=4\frac{125}{1000}-1,11=4,125-1,11=3,015\\\\2,75+\frac{2}{7}=2\frac{75}{100}+\frac{2}{7}=2\frac{3}{4}+\frac{2}{7}=2\frac{21}{28}+\frac{8}{28}=2\frac{29}{28}=3\frac{1}{28}[/tex]
Od d) do f)
[tex]0,125-\frac{1}{9}=\frac{125}{1000}-\frac{1}{9}=\frac{1}{8}-\frac{1}{9}=\frac{9}{72}-\frac{8}{72}=\frac{1}{72}\\\\3\frac{1}{3}-3,3=3\frac{1}{3}-3\frac{3}{10}=3\frac{10}{30}-3\frac{9}{30}=\frac{1}{30}\\\\1\frac{1}{5}-0,09=1\frac{2}{10}-0,09=1,2-0,09=1,11[/tex]
Pozdrawiam!
