ГДЗ по математике 6 класс Ерина Никольский рабочая тетрадь Часть 1, 2 Часть 1 | Страница 87

Авторы:
Тип:рабочая тетрадь
Часть:1, 2
Серия:Учебно-методический комплект

Страница 87

\[\mathbf{Страница\ 87.}\]

\[\boxed{\mathbf{12.}}\]

\[\textbf{а)} - \frac{11}{210} - \frac{3}{140} =\]

\[= - \frac{22}{420} - \frac{9}{420} = - \frac{31}{420}\]

\[- \frac{7}{200} + \frac{9}{150} = - \frac{21}{600} + \frac{36}{600} =\]

\[= \frac{15}{600} = \frac{1}{40}\]

\[- \frac{7}{400} + \frac{11}{350} =\]

\[= - \frac{49}{2800} + \frac{88}{2800} = \frac{39}{2800}\]

\[\textbf{б)} - \frac{7}{168} - \frac{5}{126} =\]

\[= - \frac{21}{504} - \frac{20}{504} = - \frac{41}{504}\]

\[- \frac{11}{120} + \frac{13}{90} = - \frac{33}{360} + \frac{52}{360} =\]

\[= \frac{19}{360}\]

\[- \frac{17}{240} + \frac{19}{160} = - \frac{34}{480} + \frac{57}{480} =\]

\[= \frac{23}{480}\]

\[\boxed{\mathbf{13.}}\]

\[\textbf{а)}\ - \frac{1}{7} - \frac{5}{14} - \frac{6}{21} =\]

\[= - \frac{6}{42} - \frac{15}{42} - \frac{12}{42} = - \frac{33}{42}\]

\[\frac{29}{36} - \frac{7}{9} - \frac{3}{4} = \frac{29}{36} - \frac{28}{36} - \frac{27}{36} =\]

\[= - \frac{26}{36} = - \frac{23}{18}\]

\[\frac{11}{24} - \frac{5}{16} + \frac{7}{8} = \frac{22}{48} - \frac{15}{48} + \frac{42}{48} =\]

\[= \frac{49}{48} = 1\frac{1}{48}\]

\[\textbf{б)} - \frac{3}{8} - \frac{13}{16} - \frac{11}{24} =\]

\[= - \frac{18}{48} - \frac{39}{48} - \frac{22}{48} =\]

\[= - \frac{79}{48} = - 1\frac{31}{48}\]

\[\frac{13}{45} - \frac{9}{40} - \frac{2}{5} =\]

\[= \frac{104}{360} - \frac{81}{360} - \frac{144}{360} = - \frac{121}{360}\]

\[\frac{19}{36} - \frac{7}{30} + \frac{5}{6} =\]

\[= \frac{95}{180} - \frac{42}{180} + \frac{150}{180} =\]

\[= \frac{203}{180} = 1\frac{23}{180}\]

\[\boxed{\mathbf{14.}}\]

\[\textbf{а)}\ \frac{3}{19} + x = - \frac{11}{19}\]

\[x = - \frac{11}{19} - \frac{3}{19}\]

\[x = - \frac{14}{19}.\]

\[Ответ:\ - \frac{14}{19}.\]

\[\textbf{б)}\ \frac{3}{11} - x = - \frac{1}{4}\]

\[x = \frac{3}{11} - \left( - \frac{1}{4} \right)\]

\[x = \frac{12}{44} + \frac{11}{44}\]

\[x = \frac{23}{44}.\]

\[Ответ:\ \frac{23}{44}.\]

\[\textbf{в)}\ x - \frac{1}{8} = - \frac{3}{4}\]

\[x = - \frac{6}{8} + \frac{1}{8}\]

\[x = - \frac{5}{8}.\]

\[Ответ:\ - \frac{5}{8}.\]

\[\textbf{г)}\ \frac{3}{56} - x = - \frac{4}{7}\]

\[x = \frac{3}{56} - \left( - \frac{4}{7} \right)\]

\[x = \frac{3}{56} + \frac{32}{56}\]

\[x = \frac{35}{56} = \frac{5}{8}.\]

\[Ответ:\frac{5}{8}.\]

\[\boxed{\mathbf{15.}}\]

\[\textbf{а)}\ \frac{5}{7} + \frac{3}{7} = \frac{8}{7} = 1\frac{1}{7}\]

\[- \frac{7}{18} + \frac{5}{9} = - \frac{7}{18} + \frac{10}{18} = \frac{3}{18} = \frac{1}{6}\]

\[\textbf{б)}\ \frac{6}{11} + \frac{9}{11} = \frac{15}{11} = 1\frac{4}{11}\]

\[- \frac{3}{16} + \frac{1}{8} = - \frac{3}{16} + \frac{2}{16} = - \frac{1}{16}\]

\[\boxed{\mathbf{16.}}\]

\[\textbf{а)}\ \left( 3\frac{5}{12} + 12\frac{7}{11} \right) - 1\frac{7}{11} =\]

\[= 3\frac{5}{12} + 12\frac{7}{11} - 1\frac{7}{11} =\]

\[= 3\frac{5}{12} + 11 = 14\frac{5}{12}\]

\[\textbf{б)}\ \left( \frac{5}{13} + \frac{11}{26} \right) \cdot 52 =\]

\[= \frac{5}{13} \cdot 52 + \frac{11}{26} \cdot 52 =\]

\[= 20 + 22 = 42\]

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