Решебник самостоятельные и по математике 4 класс Петерсон контрольные работы Вариант 2 Самостоятельная работа к урокам П-3

\[\boxed{\mathbf{1.}}\]

\[\boxed{\mathbf{2.}}\]

\[\frac{7}{9} > \frac{4}{9}\]

\[\frac{8}{11} > \frac{8}{15}\]

\[\frac{5}{7} < \frac{7}{5}\]

\[2\frac{1}{3} > 1\frac{2}{3}\]

\[a - \frac{3}{4} < a - \frac{1}{4}\]

\[\frac{6}{15} - n < \frac{6}{10} - n\]

\[\boxed{\mathbf{3.}}\]

\[\textbf{а)}\ \frac{15}{7} = 2\frac{1}{7}\]

\[\frac{103}{24} = 4\frac{7}{24}\]

\[\textbf{б)}\ 1\frac{3}{5} = \frac{8}{5}\]

\[3\frac{9}{16} = \frac{57}{16}\]

\[\boxed{\mathbf{4.}}\]

\[8 - \left( x + 2\frac{9}{12} \right) = 3\frac{10}{12}\]

\[x + 2\frac{9}{12} = 8 - 3\frac{10}{12}\]

\[x + 2\frac{9}{12} = 7\frac{12}{12} - 3\frac{10}{12}\]

\[x + 2\frac{9}{12} = 4\frac{2}{12}\]

\[x = 4\frac{2}{12} - 2\frac{9}{12}\]

\[x = 3\frac{14}{12} - 2\frac{9}{12}\]

\[x = 1\frac{5}{12}.\]

\[\boxed{\mathbf{5.}}\]

\[1)\ 7\frac{4}{20} - 2\frac{15}{20} = 6\frac{24}{20} - 2\frac{15}{20} =\]

\[= 4\frac{9}{20}\ (дм) - вторая\ сторона\ \]

\[треугольника.\]

\[2)\ 17 - \left( 7\frac{4}{20} + 4\frac{9}{20} \right) =\]

\[= 16\frac{20}{20} - 11\frac{13}{20} = 5\frac{7}{20}\ (дм) -\]

\[третья\ сторона.\]

\[Ответ:5\frac{7}{20}\ дм.\]

\[\boxed{\mathbf{6.}}\]

\[\frac{1}{3} = \frac{5}{15}\]