Nowa MATeMAtyka 1. Podręcznik. Zakres podstawowy | Strona 18
Zadanie 4. Oblicz.
a)
\[ -\frac{2\frac{1}{6}}{1\frac{4}{9}} = \frac{13}{6} \cdot \frac{9}{13} = \frac{9}{6} = 1\frac{1}{2}\]
b)
\[ \frac{\frac{1}{2} + \frac{1}{3}}{\frac{1}{2}} = \frac{\frac{3}{6} + \frac{2}{6}}{\frac{1}{2}} = \frac{\frac{5}{6}}{\frac{1}{2}} = \frac{5}{6} \cdot \frac{2}{1} = \frac{5}{3} = 1\frac{2}{3}\]
c)
\[\frac{1\frac{5}{12}}{\frac{1}{2} – \frac{1}{3}} = \frac{\frac{17}{12}}{\frac{3}{6} – \frac{2}{6}} = \frac{\frac{17}{12}}{\frac{1}{6}} = \frac{17}{12} \cdot \frac{6}{1} = \frac{51}{6} = 8\frac{3}{6} = 8\frac{1}{2} \]
d)
\[\frac{2 – \frac{1}{3}}{2 – \frac{1}{2}} = \frac{\frac{6}{3} – \frac{1}{3}}{\frac{4}{2} – \frac{1}{2}} = \frac{\frac{5}{3}}{\frac{3}{2}} = \frac{5}{3} \cdot \frac{2}{3} = \frac{10}{9} = 1\frac{1}{9}\]
e)
\[\frac{\frac{3}{4} + \frac{1}{8}}{\frac{3}{4} – \frac{1}{8}} = \frac{\frac{6}{8} + \frac{1}{8}}{\frac{6}{8} – \frac{1}{8}} = \frac{\frac{7}{8}}{\frac{5}{8}} = \frac{7}{8} \cdot \frac{8}{5} = \frac{7}{5} = 1\frac{2}{5}\]
f)
\[\\frac{\frac{1}{2} – \frac{2}{5}}{\frac{1}{2} – 1\frac{1}{5}} = \frac{\frac{5}{10} – \frac{4}{10}}{\frac{1}{2} – \frac{6}{5}} = \frac{\frac{1}{10}}{\frac{5}{10} – \frac{12}{10}} = \frac{\frac{1}{10}}{ – \frac{7}{10}} = \frac{1}{10} \cdot ( – \frac{10}{7}) = – \frac{1}{7}\]
g)
\[\frac{2\frac{1}{4} – 3\frac{1}{2}}{\frac{1}{8} + \frac{1}{2}} = \frac{\frac{9}{4} – \frac{7}{2}}{\frac{1}{8} + \frac{4}{8}} = \frac{\frac{9}{4} – \frac{14}{4}}{\frac{5}{8}} = \frac{ – \frac{5}{4}}{\frac{5}{8}} = – \frac{5}{4} \cdot \frac{8}{5} = – \frac{1}{1} \cdot \frac{2}{1} = – 2\]
h)
\[\frac{3\frac{1}{2} – 2 \cdot \frac{3}{4}}{2\frac{1}{2} \cdot \frac{2}{5} – 2} = \frac{\frac{7}{2} – \frac{6}{4}}{\frac{5}{2} \cdot \frac{2}{5} – 2} = \frac{\frac{14}{4} – \frac{6}{4}}{1 – 2} = \frac{ – \frac{8}{4}}{ – 1} = – 2 \cdot 1 = – 2\]
Zadanie 5. Oblicz wartość podanego wyrażenia dla x = 1 1/2, y = -1/4, z = -3/8.
a)
\[\frac{1\frac{1}{2} – ( – \frac{1}{4})}{ – \frac{1}{4} – ( – \frac{3}{8})} = \frac{\frac{3}{2} + \frac{1}{4}}{ – \frac{1}{4} + \frac{3}{8}} = \frac{\frac{6}{4} + \frac{1}{4}}{ – \frac{2}{8} + \frac{3}{8}} = \frac{\frac{7}{4}}{\frac{1}{8}} = \frac{7}{4} \cdot \frac{8}{1} = \frac{7}{1} \cdot \frac{2}{1} = 14\]
b)
\[\frac{1\frac{1}{2} + ( – \frac{1}{4}) – ( – \frac{3}{8})}{ – \frac{1}{4} + ( – \frac{3}{8})} = \frac{\frac{3}{2} – \frac{1}{4} + \frac{3}{8}}{ – \frac{1}{4} – \frac{3}{8}} = \frac{\frac{12}{8} – \frac{2}{8} + \frac{3}{8}}{ – \frac{2}{8} – \frac{3}{8}} = \frac{\frac{13}{8}}{ – \frac{5}{8}} = \frac{13}{8} \cdot ( – \frac{8}{5}) = – \frac{13}{5} = – 2\frac{3}{5}\]
c)
\[\frac{1\frac{1}{2} – ( – \frac{1}{4}) + 2( – \frac{3}{8})}{1\frac{1}{2} + ( – \frac{1}{4}) + 3( – \frac{3}{8})} = \frac{\frac{3}{2} + \frac{1}{4} – \frac{6}{8}}{\frac{3}{2} – \frac{1}{4} – \frac{9}{8}} = \frac{\frac{12}{8} + \frac{2}{8} – \frac{6}{8}}{\frac{12}{8} – \frac{2}{8} – \frac{9}{8}} = \frac{\frac{8}{8}}{\frac{1}{8}} = 1 \cdot 8 = 8\]
Zadanie 6. Uporządkuj liczby a, b, c w kolejności rosnącej.
\[a = \frac{1 – \frac{1}{8} \cdot \frac{1}{2}}{\frac{7}{8} – \frac{3}{4}} = \frac{1 – \frac{1}{16}}{\frac{7}{8} – \frac{6}{8}} = \frac{1 – \frac{1}{16}}{\frac{1}{8}} = \frac{\frac{15}{16}}{\frac{1}{8}} = \frac{15}{16} \cdot 8 = \frac{15}{2} = 7\frac{1}{2}\]
\[b = 1\frac{1}{2} \cdot 4\frac{6}{11} \cdot 3\frac{2}{3} \cdot ( – \frac{1}{5}) = \frac{3}{2} \cdot \frac{50}{11} \cdot \frac{11}{3} \cdot ( – \frac{1}{5}) = \frac{1}{2} \cdot \frac{10}{1} \cdot \frac{1}{1} \cdot ( – \frac{1}{1}) = – 5\]
\[c = \frac{2}{11} \cdot \frac{3}{2} \cdot 2\frac{3}{4} \cdot ( – 1\frac{1}{3}) = \frac{2}{11} \cdot \frac{3}{2} \cdot \frac{11}{4} \cdot ( – \frac{4}{3}) = \frac{2}{1} \cdot \frac{1}{2} \cdot \frac{1}{1} \cdot ( – \frac{1}{1}) = – 1\]
b<c<a
Zadanie 7. Uzasadnij, że nie istnieje trójkąt o bokach długości x, y i z.
a)
\[x = \frac{1}{2} \cdot (\frac{2}{3} – \frac{1}{4}) = \frac{1}{2} \cdot (\frac{8}{12} – \frac{3}{12}) = \frac{1}{2} \cdot \frac{5}{12} = \frac{5}{24}\]
\[y = \frac{1}{2} – (\frac{1}{3} – \frac{1}{4}) = \frac{1}{2} – (\frac{4}{12} – \frac{3}{12}) = \frac{6}{12} – \frac{1}{12} = \frac{5}{12}\]
\[z = (\frac{1}{3} – \frac{1}{2})^2 + (\frac{1}{4} – \frac{1}{3})^2 = (\frac{2}{6} – \frac{3}{6})^2 + (\frac{3}{12} – \frac{4}{12})^2 = ( – \frac{1}{6})^2 + ( – \frac{1}{12})^2 = \frac{1}{36} + \frac{1}{144} = \frac{4}{144} + \frac{1}{144} = \frac{5}{144}\]
\[x = \frac{5}{24} = \frac{30}{144}\]
\[y = \frac{5}{12} = \frac{60}{144}\]
Aby zbudować trójkąt suma długości krótszych boków musi być większa od długości najdłuższego boku.
\[\frac{24}{144} + \frac{5}{144} < \frac{60}{144}\]
Nie istnieje trójkąt o takich bokach.
b)
\[x = \frac{\frac{1}{2} – \frac{1}{4}}{\frac{1}{2} – \frac{2}{5}} = \frac{\frac{2}{4} – \frac{1}{4}}{\frac{5}{10} – \frac{4}{10}} = \frac{\frac{1}{4}}{\frac{1}{10}} = \frac{1}{4} \cdot \frac{10}{1} = \frac{10}{4} = 2\frac{1}{2}\]
\[y = \frac{2\frac{1}{5} – \frac{1}{4}}{2\frac{1}{2} – \frac{1}{4}} = \frac{\frac{11}{5} – \frac{1}{4}}{\frac{5}{2} – \frac{1}{4}} = \frac{\frac{44}{20} – \frac{5}{20}}{\frac{10}{4} – \frac{1}{4}} = \frac{\frac{39}{20}}{\frac{9}{4}} = \frac{39}{20} \cdot \frac{4}{9} = \frac{39}{5} \cdot \frac{1}{9} = \frac{39}{45} = \frac{13}{15}\]
\[z = \frac{1\frac{1}{3} + \frac{1}{2}}{1\frac{2}{3} + \frac{1}{4}} = \frac{\frac{4}{3} + \frac{1}{2}}{\frac{5}{3} + \frac{1}{4}} = \frac{\frac{8}{6} + \frac{3}{6}}{\frac{20}{12} + \frac{3}{12}} = \frac{\frac{11}{6}}{\frac{23}{12}} = \frac{11}{6} \cdot \frac{12}{23} = \frac{11}{1} \cdot \frac{2}{17} = \frac{22}{23}\]
Aby zbudować trójkąt suma długości krótszych boków musi być większa od długości najdłuższego boku.
\[\frac{22}{23} + \frac{13}{15} = \frac{330}{345} + \frac{299}{345} = \frac{629}{345} = 1\frac{284}{345} < 2\frac{1}{2}\]
Nie istnieje trójkąt o takich bokach.
Zadanie 8. Ułamek egipski to ułamek postaci 1/n ,gdzie n jest liczbą naturalną dodatnią. Przeczytaj informację obok i przedstaw podany ułamek jako sumę różnych ułamków egipskich.
a)
Zaokrąglamy odwrotność w górę:
\[\frac{2}{11}\rightarrow \frac{11}{2} \approx 6\]
\[\frac{2}{11} – \frac{1}{6} = \frac{12}{66} – \frac{11}{66} = \frac{1}{66}\]
Zatem:
\[\frac{2}{11} = \frac{1}{66} + \frac{1}{66}\]
b)
Zaokrąglamy odwrotność w górę:
\[\frac{2}{17}\rightarrow \frac{17}{2} \approx 9\]
\[\frac{2}{17} – \frac{1}{9} = \frac{18}{153} – \frac{17}{153} = \frac{1}{153}\]
Zatem:
\[\frac{2}{17} = \frac{1}{9} + \frac{1}{153}\]
c)
Zaokrąglamy odwrotność w górę:
\[\frac{2}{31}\rightarrow \frac{31}{2} \approx 16\]
\[\frac{2}{31} – \frac{1}{16} = \frac{132}{496} – \frac{131}{496} = \frac{1}{496}\]
Zatem:
\[\frac{2}{31} = \frac{1}{16} + \frac{1}{496}\]
Zadanie 9. Podaj liczbę odwrotną do liczby x.
a)
\[x = \frac{1}{4} + \frac{2}{3} = \frac{3}{12} + \frac{8}{12} = \frac{11}{12}\Rightarrow \frac{12}{11} = 1\frac{1}{11}\]
b)
\[x = \frac{3}{4} – \frac{3}{8} = \frac{6}{8} – \frac{3}{8} = \frac{3}{8}\Rightarrow \frac{8}{3} = 2\frac{2}{3}\]
c)
\[x = \frac{5}{6} + \frac{7}{8} = \frac{40}{48} + \frac{42}{48} = \frac{82}{48}\Rightarrow \frac{48}{82} = \frac{24}{41}\]
d)
\[x = \frac{8}{15} – \frac{4}{9} = \frac{24}{45} – \frac{20}{45} = \frac{4}{45}\Rightarrow \frac{45}{4} = 11\frac{1}{4}\]
e)
\[x = \frac{1}{2} + 13 – \frac{1}{4} = \frac{2}{4} + 13 – \frac{1}{4} = 13\frac{1}{4} = \frac{53}{4}\Rightarrow \frac{4}{53}\]
f)
\[x = \frac{2}{3} – 34 + \frac{1}{6} = \frac{4}{6} – 34 + \frac{1}{6} = – 34 + \frac{5}{6} = – 33\frac{1}{6} = – \frac{199}{6}\Rightarrow – \frac{6}{199}\]
Zadanie 10. Oblicz.
a)
\[(1\frac{4}{5} – 3\frac{2}{7}) \cdot 5\frac{5}{13} = (\frac{9}{5} – \frac{23}{7}) \cdot \frac{70}{13} = (\frac{63}{35} – \frac{115}{35}) \cdot \frac{70}{13} = – \frac{52}{35} \cdot \frac{70}{13} = – \frac{4}{1} \cdot \frac{2}{1} = – 8\]
b)
\[1\frac{4}{5} – 3\frac{2}{7} \cdot 5\frac{5}{13} = \frac{9}{5} – \frac{23}{7} \cdot \frac{70}{13} = \frac{9}{5} – \frac{23}{1} \cdot \frac{10}{13} = \frac{9}{5} – \frac{230}{13} = \frac{117}{65} – \frac{1150}{65} = – \frac{1033}{65} = – 15\frac{58}{65}\]
c)
\[(\frac{2}{5} + 1\frac{2}{3}) \cdot ( – 2)^2 = (\frac{6}{15} + 1\frac{10}{15}) \cdot 4 = (\frac{6}{15} + \frac{25}{15}) \cdot 4 = \frac{31}{15} \cdot 4 = \frac{124}{15} = 8\frac{4}{15}\]
d)
\[((2\frac{3}{4} – 3\frac{1}{2}) \cdot ( – 2)^2 = (\frac{11}{4} – \frac{7}{2}) \cdot ( – 2)^2 = (\frac{11}{4} – \frac{14}{4}) \cdot ( – 2)^2 = – \frac{3}{4} \cdot ( – 2)^2 = \frac{3}{4} \cdot 4 = 3\]
e)
\[(2\frac{1}{4} + ( – \frac{1}{2})^2)^2 = (\frac{9}{4} + \frac{1}{4})^2 = (\frac{10}{4})^2 = \frac{100}{16} = 6\frac{4}{16} = 6\frac{1}{4}\]
f)
\[(\frac{3}{4} – ( – \frac{3}{4})^2)^2 = (\frac{12}{16} – \frac{9}{16})^2 = (\frac{3}{16})^2 = \frac{9}{256}\]
Zadanie 11. Oblicz.
a)
\[\frac{\frac{1}{4} – \frac{1}{3} – \frac{1}{6}}{\frac{1}{8} – \frac{1}{5} – \frac{1}{10}} = \frac{\frac{6}{24} – \frac{8}{24} – \frac{4}{24}}{\frac{5}{40} – \frac{8}{40} – \frac{4}{40}} = \frac{ – \frac{6}{24}}{ – \frac{7}{40}} = \frac{6}{24} \cdot \frac{40}{7} = \frac{1}{4} \cdot \frac{40}{7} = \frac{10}{7} = 1\frac{3}{7}\]
b)
\[\frac{\frac{9}{5}:\frac{3}{7} – 3\frac{3}{4} \cdot \frac{2}{5}}{5\frac{1}{4} \cdot \frac{3}{7}} = \frac{\frac{9}{5} \cdot \frac{7}{3} – \frac{15}{4} \cdot \frac{2}{5}}{\frac{21}{4} \cdot \frac{3}{7}} = \frac{\frac{3}{5} \cdot \frac{7}{1} – \frac{3}{2} \cdot \frac{1}{1}}{\frac{3}{4} \cdot \frac{3}{1}} = \frac{\frac{21}{5} – \frac{3}{2}}{\frac{9}{4}} = \frac{\frac{42}{10} – \frac{15}{10}}{\frac{9}{4}} = \frac{27}{10} \cdot \frac{4}{9} = \frac{3}{10} \cdot \frac{4}{1} = \frac{12}{10} = 1\frac{1}{5}\]
c)
\[\frac{1\frac{1}{2} + 2\frac{1}{4} – \frac{2}{3} \cdot 1,5}{\frac{1}{4} \cdot ( – 2)^2 – 3\frac{1}{7} \cdot \frac{7}{11}} = \frac{\frac{3}{2} + \frac{9}{4} – \frac{2}{3} \cdot \frac{3}{2}}{\frac{1}{4} \cdot 4 – \frac{22}{7} \cdot \frac{7}{11}} = \frac{\frac{3}{2} + \frac{9}{4} – 1}{1 – 2} = \frac{\frac{6}{4} + \frac{9}{4} – 1}{ – 1} = \frac{\frac{15}{4} – 1}{ – 1} = \frac{\frac{11}{4}}{ – 1} = – \frac{11}{4} = – 2\frac{3}{4}\]