Упр.313 Часть 2 ГДЗ Дорофеев Петерсон 5 класс (Математика)
- Упрости выражение и найди его значение:
1) $$3\frac{1}{2}a+\frac{5}{6}+2\frac{5}{8}a+1\frac{3}{4}$$, если $$a=0, 1, 4, \frac{8}{49}, 1\frac{5}{7}$$;
2) $$4\frac{1}{6}b+1\frac{1}{3}+1\frac{9}{10}b+2$$, если $$b=0, 1, 5, \frac{3}{13}, 1\frac{2}{7}$$.
1) Сначала приведём подобные слагаемые:
$$3\frac{1}{2}a+\frac{5}{6}+2\frac{5}{8}a+1\frac{3}{4} = \left(3\frac{1}{2}+2\frac{5}{8}\right)a+\left(\frac{5}{6}+1\frac{3}{4}\right)$$
$$= \left(3\frac{4}{8}+2\frac{5}{8}\right)a+\left(\frac{10}{12}+1\frac{9}{12}\right) = 5\frac{1}{8}a+2\frac{7}{12}$$
Теперь подставим значения $$a$$:
При $$a=0$$:
$$5\frac{1}{8}\cdot 0+2\frac{7}{12}=2\frac{7}{12}$$
При $$a=1$$:
$$5\frac{1}{8}\cdot 1+2\frac{7}{12} = 5\frac{1}{8}+2\frac{7}{12} = 7\frac{29}{24} = 8\frac{5}{24}$$
При $$a=4$$:
$$5\frac{1}{8}\cdot 4+2\frac{7}{12} = 20\frac{1}{2}+2\frac{7}{12} = 22\frac{5}{12}$$
При $$a=\frac{8}{49}$$:
$$5\frac{1}{8}\cdot \frac{8}{49}+2\frac{7}{12} = \frac{41}{8}\cdot \frac{8}{49}+2\frac{7}{12} = \frac{41}{49}+2\frac{7}{12} = 3\frac{7}{12}$$
При $$a=1\frac{5}{7}$$:
$$5\frac{1}{8}\cdot 1\frac{5}{7}+2\frac{7}{12} = \frac{41}{8}\cdot \frac{12}{7}+2\frac{7}{12} = \frac{123}{14}+2\frac{7}{12} = 10\frac{1}{2}+2\frac{7}{12} = 13\frac{1}{12}$$
2) Аналогично упростим выражение:
$$4\frac{1}{6}b+1\frac{1}{3}+1\frac{9}{10}b+2 = \left(4\frac{1}{6}+1\frac{9}{10}\right)b+\left(1\frac{1}{3}+2\right)$$
$$= \left(4\frac{5}{30}+1\frac{27}{30}\right)b+3\frac{1}{3} = 6\frac{1}{15}b+3\frac{1}{3}$$
Подставим значения $$b$$:
При $$b=0$$:
$$6\frac{1}{15}\cdot 0+3\frac{1}{3}=3\frac{1}{3}$$
При $$b=1$$:
$$6\frac{1}{15}\cdot 1+3\frac{1}{3} = 6\frac{1}{15}+3\frac{1}{3} = 9\frac{2}{5}$$
При $$b=5$$:
$$6\frac{1}{15}\cdot 5+3\frac{1}{3} = 30\frac{1}{3}+3\frac{1}{3} = 33\frac{2}{3}$$
При $$b=\frac{3}{13}$$:
$$6\frac{1}{15}\cdot \frac{3}{13}+3\frac{1}{3} = \frac{91}{15}\cdot \frac{3}{13}+3\frac{1}{3} = \frac{7}{5}+3\frac{1}{3} = 4\frac{11}{15}$$
При $$b=1\frac{2}{7}$$:
$$6\frac{1}{15}\cdot 1\frac{2}{7}+3\frac{1}{3} = \frac{91}{15}\cdot \frac{9}{7}+3\frac{1}{3} = \frac{39}{5}+3\frac{1}{3} = 11\frac{2}{15}$$
Ответ
1) $$5\frac{1}{8}a+2\frac{7}{12}$$; при $$a=0$$ — $$2\frac{7}{12}$$, при $$a=1$$ — $$8\frac{5}{24}$$, при $$a=4$$ — $$22\frac{5}{12}$$, при $$a=\frac{8}{49}$$ — $$3\frac{7}{12}$$, при $$a=1\frac{5}{7}$$ — $$13\frac{1}{12}$$.
2) $$6\frac{1}{15}b+3\frac{1}{3}$$; при $$b=0$$ — $$3\frac{1}{3}$$, при $$b=1$$ — $$9\frac{2}{5}$$, при $$b=5$$ — $$33\frac{2}{3}$$, при $$b=\frac{3}{13}$$ — $$4\frac{11}{15}$$, при $$b=1\frac{2}{7}$$ — $$11\frac{2}{15}$$.
