Упр.1094 ГДЗ Зубарева Мордкович 6 класс ФГОС (Математика)

Задача

а) (7 1/2•2 2/3-12 1/4 :7/2)/(110•3/5)+(3 3/8+2 3/4)/(24:2 2/5);
б) (3 1/4+2 1/6) :2 3/5-2/3•2 1/4+5 1/6;
в) ((4 2/5+1 1/7)•(2 8/13+3 2/3))/(144-256);
г) (5 9/11 :1 2/9-7 2/3•1 1/23)/(11/54-5/36).

Подробный ответ

а) Преобразуем смешанные числа в неправильные дроби и выполняем действия по порядку:

$$ \frac{7\frac{1}{2}\cdot 2\frac{2}{3}-12\frac{1}{4}: \frac{7}{2}}{110\cdot \frac{3}{5}}+\frac{3\frac{3}{8}+2\frac{3}{4}}{24:2\frac{2}{5}} = \frac{\frac{15}{2}\cdot \frac{8}{3}-\frac{49}{4}:\frac{7}{2}}{22\cdot 3} + \frac{\frac{3}{8}+\frac{2}{8}}{24:\frac{12}{5}} $$

$$ = \frac{5\cdot 4-\frac{7}{2}}{66} + \frac{5\frac{5}{8}}{24\cdot \frac{5}{12}} = \frac{20-3\frac{1}{2}}{66} + \frac{\frac{49}{8}}{10} $$

$$ = \frac{16\frac{1}{2}}{66}+\frac{49}{8}\cdot \frac{1}{10} = \frac{33}{132}+\frac{49}{80} = \frac{1}{4}+\frac{49}{80} = \frac{20}{80}+\frac{49}{80} = \frac{69}{80} $$

б)

$$ \left(3\frac{1}{4}+2\frac{1}{6}\right):2\frac{3}{5}-\frac{2}{3}\cdot 2\frac{1}{4}+5\frac{1}{6} = \left(3\frac{1}{12}+2\frac{2}{12}\right):\frac{13}{5}-\frac{2}{3}\cdot \frac{9}{4}+\frac{31}{6} $$

$$ = 5\frac{1}{12}\cdot \frac{5}{13}-\frac{3}{2}+5\frac{1}{6} = \frac{65}{12}\cdot \frac{5}{13}-\frac{3}{2}+5\frac{1}{6} $$

$$ = \frac{25}{12}-\frac{3}{2}+5\frac{1}{6} = 2\frac{1}{12}+5\frac{1}{6} = 2\frac{1}{12}+5\frac{2}{12} = 7\frac{3}{12} = 5\frac{3}{4} $$

в)

$$ \frac{\left(4\frac{2}{5}+1\frac{1}{7}\right)\cdot \left(2\frac{8}{13}+3\frac{2}{3}\right)}{144-256} = \frac{\left(4\frac{14}{35}+1\frac{5}{35}\right)\cdot \left(2\frac{24}{39}+3\frac{26}{39}\right)}{-112} $$

$$ = \frac{5\frac{19}{35}\cdot 5\frac{50}{39}}{-112} = \frac{\frac{194}{35}\cdot \frac{245}{39}}{-112} = \frac{\frac{194}{1}\cdot \frac{7}{39}}{-112} = \frac{1358}{39}:(-112) $$

$$ = -\frac{1358}{39}\cdot \frac{1}{112} = -\frac{679}{39\cdot 56} = -\frac{97}{39\cdot 8} = -\frac{97}{312} $$

г)

$$ \frac{5\frac{9}{11}:1\frac{2}{9}-7\frac{2}{3}\cdot 1\frac{1}{23}}{\frac{11}{54}-\frac{5}{36}} = \frac{\frac{64}{11}:\frac{11}{9}-\frac{23}{3}\cdot \frac{24}{23}}{\frac{22-15}{108}} $$

$$ = \frac{\frac{64}{11}\cdot \frac{9}{11}-8}{\frac{7}{108}} = \frac{\frac{576}{121}-8}{\frac{7}{108}} = \frac{-3\frac{29}{121}}{\frac{7}{108}} $$

$$ = -\frac{392}{121}:\frac{7}{108} = -\frac{392}{121}\cdot \frac{108}{7} = -\frac{6048}{121} = -49\frac{119}{121} $$