ГДЗ по математике 6 класс Ерина Виленкин рабочая тетрадь Часть 1, 2 Часть 1 | Страница 77

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

\[\textbf{а)}\frac{11}{13}x + \frac{2}{13}x =\]

\[= \left( \frac{11}{13} + \frac{2}{13} \right)x = x\]

\[\textbf{б)}\frac{7^{\backslash 2}}{8}a - \frac{9}{16}a = \left( \frac{14}{16} - \frac{9}{16} \right)a =\]

\[= \frac{5}{16}a\]

\[\textbf{в)}\ 5\frac{1}{6}b + \frac{2^{\backslash 2}}{3}b = \left( 5\frac{1}{6} + \frac{4}{6} \right)b =\]

\[= 5\frac{5}{6}b\]

\[\textbf{г)}\ 7\frac{3^{\backslash 2}}{4}y - 2\frac{7}{8}y =\]

\[= \left( 7\frac{6}{8} - 2\frac{7}{8} \right)y =\]

\[= \left( 6\frac{14}{8} - 2\frac{7}{8} \right)y = 4\frac{7}{8}y\]

\[\textbf{д)}\ 2\frac{1}{2}x - \left( 1\frac{1}{4}x - \frac{1}{2}x \right) =\]

\[= 2\frac{1}{2}x + \frac{1}{2}x - 1\frac{1}{4}x =\]

\[= \left( 2\frac{2}{2} - 1\frac{1}{4} \right)x =\]

\[= \left( 2\frac{4}{4} - 1\frac{1}{4} \right)x = 1\frac{3}{4}x\]

\[\textbf{е)}\ 3\frac{4}{5}x - \left( 2\frac{3}{4}x - 1\frac{1}{5}x \right) =\]

\[= 3\frac{4}{5}x + 1\frac{1}{5}x - 2\frac{3}{4}x =\]

\[= \left( 3\frac{4}{5} + 1\frac{1}{5} - 2\frac{3}{4} \right)x =\]

\[= \left( 5 - 2\frac{3}{4} \right)x = \left( 4\frac{4}{4} - 2\frac{3}{4} \right)x =\]

\[= 2\frac{1}{4}x\]

\[\textbf{ж)}\ 3\frac{7}{15}x - \left( 2\frac{1}{2}x - 1\frac{3}{15}x \right) =\]

\[= 3\frac{7}{15}x + 1\frac{3}{15}x - 2\frac{1}{2}x =\]

\[= \left( 3\frac{7}{15} + 1\frac{3}{15} - 2\frac{1}{2} \right)x =\]

\[= \left( 4\frac{10}{15} - 2\frac{1}{2} \right)x =\]

\[= \left( 4\frac{2^{\backslash 2}}{3} - 2\frac{1^{\backslash 3}}{2} \right)x =\]

\[= \left( 4\frac{4}{6} - 2\frac{3}{6} \right)x = 2\frac{1}{6}x\]

\[\textbf{з)}\ 4\frac{7}{32}x - \left( 3\frac{3}{4}x - 1\frac{7}{32}x \right) =\]

\[= 4\frac{7}{32}x - 3\frac{3}{4}x + 1\frac{7}{32}x =\]

\[= \left( 4\frac{7}{32} + 1\frac{7}{32} - 3\frac{3}{4} \right)x =\]

\[= \left( 5\frac{14}{32} - 3\frac{3^{\backslash 4}}{4} \right)x =\]

\[= \left( 5\frac{7}{16} - 3\frac{12}{16} \right)x =\]

\[= \left( 4\frac{23}{16} - 3\frac{12}{16} \right)x = 1\frac{11}{16}x\]