Oblicz :
a) \(4- \frac{5}{12}=\)
b) \(3 \frac{3}{4}+1 \frac{1}{2}*5 \frac{5}{9}=\)
c) \(17 \frac{4}{5}* \frac{13}{178}- \frac{2}{3}: \frac{38}{39}=\)
d) \(( \frac{1}{2})^3- \frac{4}{3}+ \frac{4}{3}* \frac{2}{11}=\)
e) \(( \frac{2}{3})^2:( 1\frac{1}{2})^2+3 \frac{1}{5}-2=\)
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octahedron
- Expert
- Posty: 6762
- Rejestracja: 19 mar 2011, 00:22
- Otrzymane podziękowania: 3034 razy
- Płeć:
\(a)\,4-\frac{5}{12}=3\frac{12}{12}-\frac{5}{12}=3\frac{7}{12}
b)\,3\frac{3}{4}+1\frac{1}{2}\cdot 5\frac{5}{9}=3\frac{3}{4}+\frac{3}{2}\cdot\frac{49}{9}=3\frac{3}{4}+\frac{49}{6}=3\frac{3}{4}+8\frac{1}{6}=3\frac{9}{12}+8\frac{2}{12}=11\frac{11}{12}
c)\,17\frac{4}{5}\cdot\frac{13}{178}-\frac{2}{3}:\frac{38}{39}=\frac{89}{5}\cdot\frac{13}{178}-\frac{2}{3}\cdot\frac{39}{38}=\frac{13}{10}-\frac{13}{19}=\frac{13(19-10)}{190}=\frac{117}{190}
d)\,\(\frac{1}{2}\)^3-\frac{4}{3}+\frac{4}{3}\cdot\frac{2}{11}=\frac{1}{8}-\frac{44}{33}+\frac{8}{33}=\frac{1}{8}-\frac{36}{33}=\frac{1}{8}-\frac{12}{11}=\frac{11}{88}-\frac{96}{88}=-\frac{85}{88}
e)\,\(\frac{2}{3}\)^2:\(1\frac{1}{2}\)^2+3\frac{1}{5}-2=\(\frac{2}{3}\)^2:\(\frac{3}{2}\)^2+1\frac{1}{5}=\(\frac{2}{3}\)^2\cdot\(\frac{2}{3}\)^2+1\frac{1}{5}=\frac{16}{81}+1\frac{1}{5}=\frac{80}{405}+1\frac{81}{405}=1\frac{161}{405}\)
b)\,3\frac{3}{4}+1\frac{1}{2}\cdot 5\frac{5}{9}=3\frac{3}{4}+\frac{3}{2}\cdot\frac{49}{9}=3\frac{3}{4}+\frac{49}{6}=3\frac{3}{4}+8\frac{1}{6}=3\frac{9}{12}+8\frac{2}{12}=11\frac{11}{12}
c)\,17\frac{4}{5}\cdot\frac{13}{178}-\frac{2}{3}:\frac{38}{39}=\frac{89}{5}\cdot\frac{13}{178}-\frac{2}{3}\cdot\frac{39}{38}=\frac{13}{10}-\frac{13}{19}=\frac{13(19-10)}{190}=\frac{117}{190}
d)\,\(\frac{1}{2}\)^3-\frac{4}{3}+\frac{4}{3}\cdot\frac{2}{11}=\frac{1}{8}-\frac{44}{33}+\frac{8}{33}=\frac{1}{8}-\frac{36}{33}=\frac{1}{8}-\frac{12}{11}=\frac{11}{88}-\frac{96}{88}=-\frac{85}{88}
e)\,\(\frac{2}{3}\)^2:\(1\frac{1}{2}\)^2+3\frac{1}{5}-2=\(\frac{2}{3}\)^2:\(\frac{3}{2}\)^2+1\frac{1}{5}=\(\frac{2}{3}\)^2\cdot\(\frac{2}{3}\)^2+1\frac{1}{5}=\frac{16}{81}+1\frac{1}{5}=\frac{80}{405}+1\frac{81}{405}=1\frac{161}{405}\)