Oblicz.- Zadanie 4: Matematyka 6. Podręcznik cz. 2

Rozwiązanie

Wykonajmy obliczenia.

$- 2, 2 \cdot 1 \frac{13}{22} + 2, 5 \cdot 2 \frac{1}{2} = - \frac{22 ^{1}}{10} \cdot \frac{35}{22 _{1}} + 2, 5 \cdot 2, 5 = - 3, 5 + 6, 25 = 2, 75$

$3, 4 : (- \frac{2}{5}) - \frac{5}{6} \cdot 6, 6 = 3, 4 : (- 0, 4) - \frac{5}{6 _{1}} \cdot \frac{66 ^{11}}{10} = - 34 : 4 - \frac{55}{10} = - 8, 5 - 5, 5 = - 14$

$- 5, 6 \cdot (- \frac{5}{6}) \cdot (- 1 \frac{11}{28} + 2 \frac{1}{4}) = \frac{56}{10} \cdot \frac{5}{6} \cdot (- 1 \frac{11}{28} + 2 \frac{7}{28}) = \frac{56 ^{28}}{10 _{2}} \cdot \frac{5 ^{1}}{6 _{3}} \cdot \frac{6}{7} = \frac{28 ^{4}}{2} \cdot \frac{1}{3} \cdot \frac{6}{7 _{1}} = \frac{24}{6} = 4$

$- 2, 2 \cdot (3 \frac{1}{3} - 5 \frac{11}{15}) = - \frac{22}{10} \cdot (3 \frac{5}{15} - 5 \frac{11}{15}) = - \frac{22}{10} \cdot (- 2 \frac{6}{15}) = \frac{22}{10 _{5}} \cdot \frac{36 ^{18}}{15} = \frac{396}{75} = \frac{132}{25} = 5 \frac{7}{25}$

$0, 4 \cdot (- 1 \frac{2}{5}) : 0, 1 \cdot (- \frac{5}{14}) = 0, 4 \cdot 1, 4 : 0, 1 \cdot \frac{5}{14} = 0, 56 : 0, 1 \cdot \frac{5}{14} = 5, 6 \cdot \frac{5}{14} = \frac{56 ^{4}}{10} \cdot \frac{5}{14 _{1}} = \frac{20}{10} = 2$

$- \frac{2}{3} : 1, 6 \cdot (- 1, 5) : (- 1 \frac{7}{8}) = - \frac{2}{3} : \frac{16}{10} \cdot \frac{3}{2} : \frac{15}{8} = - \frac{2 ^{1}}{3} \cdot \frac{10}{16 _{8}} \cdot \frac{3 ^{1}}{2} \cdot \frac{8}{15 _{5}} = - \frac{1}{3} \cdot \frac{10}{8} \cdot \frac{1}{2 _{1}} \cdot \frac{8 ^{4}}{5} = - \frac{1}{3} \cdot \frac{10}{8 _{2}} \cdot \frac{4 ^{1}}{5} = - \frac{10}{60} = - \frac{1}{6}$

$\frac{5}{6} + (- 1, 5) - \frac{1}{12} + 0, 75 = \frac{5}{6} - 1 \frac{1}{2} - \frac{1}{12} + \frac{3}{4} = \frac{10}{12} - 1 \frac{6}{12} - \frac{1}{12} + \frac{9}{12} = 0$

$- 0, 1 - (- 3 \frac{1}{5}) + (- 5, 4) = - 0, 1 + 3, 2 + (- 5, 4) = - 2, 3$

$1 \frac{1}{6} - (- 0, 5)^{3} + (- \frac{1}{4})^{2} = 1 \frac{1}{6} - (- 0, 125) + \frac{1}{16} = 1 \frac{1}{6} + \frac{125}{1000} + \frac{1}{16} = 1 \frac{1}{6} + \frac{1}{8} + \frac{1}{16} = 1 \frac{8}{48} + \frac{6}{48} + \frac{3}{48} = 1 \frac{17}{48}$

$- 2 \frac{2}{3} : (- \frac{2}{3})^{2} + (- 2 \frac{5}{6}) = - \frac{8}{3} : \frac{4}{9} - 2 \frac{5}{6} = - \frac{8 ^{2}}{3 _{1}} \cdot \frac{9 ^{3}}{4 _{1}} - 2 \frac{5}{6} = - 6 - 2 \frac{5}{6} = - 8 \frac{5}{6}$

$(1, 6 - 1 \frac{3}{4})^{2} \cdot 4 \frac{4}{9} = (1, 6 - 1, 75)^{2} \cdot \frac{40}{9} = (- 0, 15)^{2} \cdot \frac{40}{9} = 0, 0225 \cdot \frac{40}{9} = \frac{225 ^{25}}{10000} \cdot \frac{40}{9 _{1}} = \frac{1000}{10000} = 0, 1$

$(- \frac{2}{3})^{2} - (- 1, 5)^{3} = \frac{4}{9} - (- \frac{3}{2})^{3} = \frac{4}{9} - (- \frac{27}{8}) = \frac{4}{9} + \frac{27}{8} = \frac{4}{9} + 3 \frac{3}{8} = \frac{32}{72} + 3 \frac{27}{72} = 3 \frac{59}{72}$

$\frac{1 , 5 + ( - 1 \frac{3}{5} )}{3 , 3 + ( - 3 \frac{1}{3} )} = \frac{1 , 5 - 1 , 6}{3 \frac{3}{10} - 3 \frac{1}{3}} = \frac{- 0 , 1}{3 \frac{1}{30} - 3 \frac{9}{30}} = \frac{\frac{1}{10}}{\frac{1}{30}} = \frac{1}{10} \cdot \frac{30}{1} = 3$

$(- 1 \frac{1}{3} - 0, 3) \cdot (- 8 \frac{4}{7}) = (- 1 \frac{1}{3} - \frac{3}{10}) \cdot (- \frac{60}{7}) = (- 1 \frac{10}{30} - \frac{9}{30}) \cdot (- \frac{60}{7}) = (- 1 \frac{19}{30}) \cdot (- \frac{60}{7}) = \frac{49 ^{7}}{30 _{1}} \cdot \frac{60 ^{2}}{7 _{1}} = 14$

$(5, 125 + (- 6 \frac{1}{2})) \cdot (- 1 \frac{3}{4} + 0, 75) = (5, 125 - 6, 5) \cdot (- 1, 75 + 0, 75) = (- 1, 375) \cdot (- 1) = 1, 375$

$\frac{- 5 \frac{5}{8} + 2 , 25}{0 , 8 \cdot 1 \frac{1}{8}} = \frac{- 5 , 625 + 2 , 25}{\frac{8 ^{1}}{10} \cdot \frac{9}{8 _{1}}} = \frac{- 3 , 375}{\frac{9}{10}} = - 3, 375 : 0, 9 = - 33, 75 : 9 = - 3, 75$