Номер 2.463, страница 105, часть 1 - гдз по математике 6 класс учебник Виленкин, Жохов

Часть 1. Параграф 2. Действия со смешенными числами. 15. Действие деления смешанных чисел - номер 2.463, страница 105.

Условие.

2.463. Найдите значение выражения:

а) 45 : 113 + 213 · 37 – 1 : 138 ;

б) 216 : (1115 – 15) + (2 18 + 34) : 534;

в) (14 + 11114) · 1457 – 23 : 116 · 732;

г) (212 : 313 + 313 : 212) · 935;

д) (10513 – 72326) : 56;

е) ((112)³ – 34) : 78.

Решение 1.

2.463

$$\begin{gathered} \mathit{а}\mathbf{)} \frac{4}{5}\stackrel{1}{ :} 1\frac{1}{3} \stackrel{4}{+} 2\frac{1}{3} \stackrel{2}{·} \frac{3}{7} \stackrel{5}{-} 1\stackrel{3}{ : }1\frac{3}{8} = \frac{48}{55}; \\ 1) \frac{4}{5} : 1\frac{1}{3} = \frac{4}{5} : \frac{4}{3} = \frac{\cancel{4}}{5} · \frac{3}{\cancel{4}} = \frac{3}{5}; \\ 2) 2\frac{1}{3} · \frac{3}{7} = \frac{7}{3} · \frac{3}{7} = 1; \\ 3) 1 : 1\frac{3}{8} = 1 : \frac{11}{8} = 1 · \frac{8}{11} = \frac{8}{11}; \\ 4) \frac{3}{5} + 1 =1\frac{3}{5}; \\ 5) 1\frac{3}{5} - \frac{8}{11}={\frac{8}{5}}^{\overline{)·11}}- {\frac{8}{11}}^{\overline{)·5}}=\frac{88}{55} - \frac{40}{55} = \frac{48}{55}. \end{gathered}$$

$$\begin{gathered} \mathit{б}\mathbf{)}\mathbf{ }2\frac{1}{6} \stackrel{3}{:} \left(1 \frac{1}{15} \stackrel{1}{-} \frac{1}{5}\right) \stackrel{5}{+} \left(2\frac{1}{8} \stackrel{2}{+} \frac{3}{4}\right) \stackrel{4}{:} 5\frac{3}{4} = 3 \\ 1) 1 \frac{1}{15} - {\frac{1}{5}}^{\overline{)·3}} = \frac{16}{15} - \frac{3}{15} = \frac{13}{15}; \\ 2) 2\frac{1}{8} + {\frac{3}{4}}^{\overline{)·2}} = 2\frac{1}{8} + \frac{6}{8} = 2\frac{7}{8}; \\ 3) 2\frac{1}{6} : \frac{13}{15} = \frac{13}{6} : \frac{13}{15} = \frac{\cancel{13}}{{\displaystyle \underset{2}{\bcancel{6}}}} · \frac{{\displaystyle \stackrel{5}{\bcancel{15}}}}{\cancel{13}} = \\ = \frac{1 · 5 }{2 · 1 }= \frac{5}{2}; \\ 4) 2 \frac{7}{8} : 5\frac{3}{4} = \frac{23}{8} : \frac{23}{4} = \frac{\cancel{23}}{{\displaystyle \underset{2}{\bcancel{8}}}} · \frac{{\displaystyle \stackrel{1}{\bcancel{4}}}}{\cancel{23}}= \\ = \frac{1}{2}; \\ 5) 2\frac{1}{2} + \frac{1}{2} = 3. \end{gathered}$$

$$\begin{gathered} \mathit{в}\mathbf{)}\mathbf{ }\left(\frac{1}{4} \stackrel{1}{+} 1\frac{11}{14}\right) \stackrel{2}{·} \frac{14}{57} \stackrel{5}{-} \frac{2}{3}\stackrel{3}{ :} 1\frac{1}{6} \stackrel{4}{· }\frac{7}{32} = \frac{3}{8} \\ 1) {\frac{1}{4}}^{\overline{)· 7}} + {1\frac{11}{14}}^{\overline{)·2}} = \frac{7}{28} + 1\frac{22}{28} = 1 + \\ + 1\frac{1}{28} = 2\frac{1}{28}; \\ 2) 2\frac{1}{28} · \frac{14}{57} = \frac{\cancel{57}}{{\displaystyle \underset{2}{\bcancel{28}}}} · \frac{{\displaystyle \stackrel{1}{\bcancel{14}}}}{\cancel{57}} = \frac{1 · 1}{2 · 1} = \frac{1}{2}; \\ 3) \frac{2}{3} : 1\frac{1}{6} = \frac{2}{3} : \frac{7}{6} = \frac{2}{{\displaystyle \underset{1}{\cancel{3}}}} · \frac{{\displaystyle \stackrel{2}{\cancel{6}}}}{7} = \frac{2 · 2 }{1 · 7} = \frac{4}{7}; \\ 4) \frac{{\displaystyle \stackrel{1}{\bcancel{4}}}}{\cancel{7}} · \frac{\cancel{7}}{{\displaystyle \underset{8}{\bcancel{32}}}} = \frac{1}{8}; \\ 5) {\frac{1}{2}}^{\overline{)·4}} - \frac{1}{8} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8}. \end{gathered}$$

$$\begin{gathered} \mathit{г}\mathbf{)} \left(2\frac{1}{2}\stackrel{1}{ : }3\frac{1}{3} \stackrel{3}{+} 3\frac{1}{3}\stackrel{2}{ :} 2\frac{1}{2}\right) \stackrel{4}{·} 9\frac{3}{5} = 20 \\ 1) 2\frac{1}{2}: 3\frac{1}{3} = \frac{5}{2} : \frac{10}{3} = \frac{{\displaystyle \stackrel{1}{\bcancel{5}}}}{2} · \frac{3}{{\displaystyle \underset{2}{\bcancel{10}}}} = \\ =\frac{1 · 3 }{2 · 2} = \frac{3}{4}; \\ 2) 3\frac{1}{3} : 2\frac{1}{2} = \frac{10}{3} : \frac{5}{2} = \frac{{\displaystyle \stackrel{2}{\cancel{10}}}}{3} · \frac{2}{{\displaystyle \underset{1}{\cancel{5}}}} = \\ = \frac{2 · 2}{3 · 1} = \frac{4}{3} = 1\frac{1}{3}; \\ 3) {\frac{3}{4}}^{\overline{)·3}} + {1\frac{1}{3}}^{\overline{)·4}} = \frac{9}{12} + 1\frac{4}{12} = 1\frac{13}{12}= \\ =1 + 1\frac{1}{12} = 2\frac{1}{12}; \\ 4) 2\frac{1}{12} · 9 \frac{3}{5} = \frac{{\displaystyle \stackrel{5}{\bcancel{25}}}}{{\displaystyle \underset{1}{\cancel{12}}}} · \frac{{\displaystyle \stackrel{4}{\cancel{48}}}}{{\displaystyle \underset{1}{\bcancel{5}}}}=\frac{5 · 4}{1 · 1} = 20. \end{gathered}$$

$$\begin{gathered} \mathit{д}\mathbf{)}\mathbf{ }\left(10 \frac{5}{13} \stackrel{1}{-} 7 \frac{23}{26}\right) \stackrel{2}{:} \frac{5}{6} =3 \\ 1) {10 \frac{5}{13}}^{\overline{)·2}} - 7 \frac{23}{26}=10 \frac{10}{26} - 7 \frac{23}{26}= \\ =9\frac{36}{25} - 7 \frac{23}{26}=2\frac{{\displaystyle \stackrel{1}{\cancel{13}}}}{{\displaystyle \underset{2}{\cancel{26}}}} = 2\frac{1}{2}; \\ 2) 2\frac{1}{2} : \frac{5}{6} = \frac{\cancel{5}}{2} · \frac{6}{\cancel{5}} =\frac{6}{2} =3. \end{gathered}$$

$$\begin{gathered} \mathit{е}\mathbf{)} \left({\left(1\frac{1}{2}\right)}^{\stackrel{1}{3}}\stackrel{2}{-}\frac{3}{4}\right) \stackrel{3}{:} \frac{7}{8} =3 \\ 1){\left(1\frac{1}{2}\right)}^3 = {\left(\frac{3}{2}\right)}^3=\frac{3}{2} · \frac{3}{2} · \frac{3}{2} = \frac{27}{8}; \\ 2) \frac{27}{8} - {\frac{3}{4}}^{\overline{)·2}}=\frac{27}{8} - \frac{6}{8}=\frac{21}{8}; \\ 3) \frac{21}{8} : \frac{7}{8} = \frac{21}{\cancel{8}} · \frac{\cancel{8}}{7} =\frac{21}{7} = 3. \end{gathered}$$

Решение 2.

а) $\frac{4}{5} : 1\frac{1}{3} + 2\frac{1}{3} \cdot \frac{3}{7} - 1 : 1\frac{3}{8} = \frac{4}{5} : \frac{4}{3} + \frac{7}{3} \cdot \frac{3}{7} - 1 : \frac{11}{8} = \frac{4}{5} \cdot \frac{3}{4} + \frac{7 \cdot 3}{3 \cdot 7} - 1 \cdot \frac{8}{11} = \frac{3}{5} + 1 - \frac{8}{11} = \frac{8}{5} - \frac{8}{11} = \frac{88 - 40}{55} = \frac{48}{55}$.
Ответ: $\frac{48}{55}$.

б) $2\frac{1}{6} : (1\frac{1}{15} - \frac{1}{5}) + (2\frac{1}{8} + \frac{3}{4}) : 5\frac{3}{4} = \frac{13}{6} : (\frac{16}{15} - \frac{1 \cdot 3}{5 \cdot 3}) + (\frac{17}{8} + \frac{3 \cdot 2}{4 \cdot 2}) : \frac{23}{4} = \frac{13}{6} : \frac{13}{15} + \frac{23}{8} : \frac{23}{4} = \frac{13}{6} \cdot \frac{15}{13} + \frac{23}{8} \cdot \frac{4}{23} = \frac{15}{6} + \frac{4}{8} = \frac{5}{2} + \frac{1}{2} = \frac{6}{2} = 3$.
Ответ: 3.

в) $(\frac{1}{4} + 1\frac{11}{14}) \cdot \frac{14}{57} - \frac{2}{3} : 1\frac{1}{6} \cdot \frac{7}{32} = (\frac{1}{4} + \frac{25}{14}) \cdot \frac{14}{57} - \frac{2}{3} : \frac{7}{6} \cdot \frac{7}{32} = (\frac{7}{28} + \frac{50}{28}) \cdot \frac{14}{57} - (\frac{2}{3} \cdot \frac{6}{7}) \cdot \frac{7}{32} = \frac{57}{28} \cdot \frac{14}{57} - \frac{4}{7} \cdot \frac{7}{32} = \frac{14}{28} - \frac{4}{32} = \frac{1}{2} - \frac{1}{8} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8}$.
Ответ: $\frac{3}{8}$.

г) $(2\frac{1}{2} : 3\frac{1}{3} + 3\frac{1}{3} : 2\frac{1}{2}) \cdot 9\frac{3}{5} = (\frac{5}{2} : \frac{10}{3} + \frac{10}{3} : \frac{5}{2}) \cdot \frac{48}{5} = (\frac{5}{2} \cdot \frac{3}{10} + \frac{10}{3} \cdot \frac{2}{5}) \cdot \frac{48}{5} = (\frac{3}{4} + \frac{4}{3}) \cdot \frac{48}{5} = (\frac{9 + 16}{12}) \cdot \frac{48}{5} = \frac{25}{12} \cdot \frac{48}{5} = 5 \cdot 4 = 20$.
Ответ: 20.

д) $(10\frac{5}{13} - 7\frac{23}{26}) : \frac{5}{6} = (\frac{135}{13} - \frac{205}{26}) : \frac{5}{6} = (\frac{270}{26} - \frac{205}{26}) : \frac{5}{6} = \frac{65}{26} : \frac{5}{6} = \frac{5}{2} \cdot \frac{6}{5} = \frac{6}{2} = 3$.
Ответ: 3.

е) $((1\frac{1}{2})^3 - \frac{3}{4}) : \frac{7}{8} = ((\frac{3}{2})^3 - \frac{3}{4}) : \frac{7}{8} = (\frac{27}{8} - \frac{6}{8}) : \frac{7}{8} = \frac{21}{8} : \frac{7}{8} = \frac{21}{8} \cdot \frac{8}{7} = \frac{21}{7} = 3$.
Ответ: 3.

Решение 3.

Решение 4.