[tex]6,5:(2\frac{1}{4}\cdot6-0,5)=\frac{65}{10}:(\frac{9}{\not4_{2}}\cdot\not6^3-0,5)=\frac{13}{2}:(\frac{9\cdot3}{2}-0,5)=\frac{13}{2}:(\frac{27}{2}-\frac{5}{10})=\\\\=\frac{13}{2}:(13\frac{1}{2}-\frac{1}{2})=\frac{13}{2}:13=\frac{\not13^1}{2}\cdot\frac{1}{\not13_{1}}=\frac{1}{2}[/tex]
[tex](5-1\frac{1}{2}):(2\frac{3}{8}-1,75)=(4\frac{2}{2}-1\frac{1}{2}):(2\frac{3}{8}-\frac{175}{100})=3\frac{1}{2}:(\frac{19}{8}-\frac{7}{4})=3\frac{1}{2}:(\frac{19}{8}-\frac{14}{8})=\\\\=3\frac{1}{2}:\frac{5}{8}=\frac{7}{\not2_{1}}\cdot\frac{\not8^4}{5}=\frac{28}{5}=5\frac{3}{5}=5\frac{6}{10}=5,6[/tex]
[tex]3\frac{3}{4}\cdot0,8+2\frac{2}{3}\cdot0,3=\frac{15}{4}\cdot\frac{8}{10}+\frac{\not8^4}{\not3_{1}}\cdot\frac{\not3^1}{\not10_{5}}=\frac{\not15^3}{\not4_{1}}\cdot\frac{\not4^1}{\not5_{1}}+\frac{4}{5}=3+\frac{4}{5}=3\frac{4}{5}=3\frac{8}{10}=3,8[/tex]
[tex]\frac{5}{7}\cdot4,2-(1\frac{1}{4}-0,5)=\frac{5}{7}\cdot\frac{42}{10}-(\frac{5}{4}-\frac{5}{10})=\frac{\not5^1}{\not7_{1}}\cdot\frac{\not21^3}{\not5_{1}}-(\frac{5}{4}-\frac{1}{2})=3-(\frac{5}{4}-\frac{2}{4})=\\\\=3-\frac{3}{4}=2\frac{4}{4}-\frac{3}{4}=2\frac{1}{4}[/tex]
[tex]4\frac{1}{2}-2,4:2\frac{2}{3}+1,4=4\frac{1}{2}-\frac{24}{10}:\frac{8}{3}+1,4=\frac{9}{2}-\frac{\not12^3}{5}\cdot\frac{3}{\not8_{2}}+\frac{14}{10}=\frac{9}{2}-\frac{9}{10}+\frac{7}{5}=\\\\=\frac{45}{10}-\frac{9}{10}+\frac{14}{10}=\frac{59}{10}-\frac{9}{10}= \frac{50}{10}=5[/tex]
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[tex]6,5:(2\frac{1}{4}\cdot6-0,5)=\frac{65}{10}:(\frac{9}{\not4_{2}}\cdot\not6^3-0,5)=\frac{13}{2}:(\frac{9\cdot3}{2}-0,5)=\frac{13}{2}:(\frac{27}{2}-\frac{5}{10})=\\\\=\frac{13}{2}:(13\frac{1}{2}-\frac{1}{2})=\frac{13}{2}:13=\frac{\not13^1}{2}\cdot\frac{1}{\not13_{1}}=\frac{1}{2}[/tex]
[tex](5-1\frac{1}{2}):(2\frac{3}{8}-1,75)=(4\frac{2}{2}-1\frac{1}{2}):(2\frac{3}{8}-\frac{175}{100})=3\frac{1}{2}:(\frac{19}{8}-\frac{7}{4})=3\frac{1}{2}:(\frac{19}{8}-\frac{14}{8})=\\\\=3\frac{1}{2}:\frac{5}{8}=\frac{7}{\not2_{1}}\cdot\frac{\not8^4}{5}=\frac{28}{5}=5\frac{3}{5}=5\frac{6}{10}=5,6[/tex]
[tex]3\frac{3}{4}\cdot0,8+2\frac{2}{3}\cdot0,3=\frac{15}{4}\cdot\frac{8}{10}+\frac{\not8^4}{\not3_{1}}\cdot\frac{\not3^1}{\not10_{5}}=\frac{\not15^3}{\not4_{1}}\cdot\frac{\not4^1}{\not5_{1}}+\frac{4}{5}=3+\frac{4}{5}=3\frac{4}{5}=3\frac{8}{10}=3,8[/tex]
[tex]\frac{5}{7}\cdot4,2-(1\frac{1}{4}-0,5)=\frac{5}{7}\cdot\frac{42}{10}-(\frac{5}{4}-\frac{5}{10})=\frac{\not5^1}{\not7_{1}}\cdot\frac{\not21^3}{\not5_{1}}-(\frac{5}{4}-\frac{1}{2})=3-(\frac{5}{4}-\frac{2}{4})=\\\\=3-\frac{3}{4}=2\frac{4}{4}-\frac{3}{4}=2\frac{1}{4}[/tex]
[tex]4\frac{1}{2}-2,4:2\frac{2}{3}+1,4=4\frac{1}{2}-\frac{24}{10}:\frac{8}{3}+1,4=\frac{9}{2}-\frac{\not12^3}{5}\cdot\frac{3}{\not8_{2}}+\frac{14}{10}=\frac{9}{2}-\frac{9}{10}+\frac{7}{5}=\\\\=\frac{45}{10}-\frac{9}{10}+\frac{14}{10}=\frac{59}{10}-\frac{9}{10}= \frac{50}{10}=5[/tex]
Kolejność wykonywania działań
- działania w nawiasach
- mnożenie lub dzielenie w kolejności ich występowania
- dodawanie lub odejmowanie w kolejności ich występowania