Упр.199 Вариант 3 Дидактические материалы ГДЗ Мерзляк Полонский 6 класс (Математика)

Найдите значение выражения:

1. $$\left(-\frac{4}{15}+\frac{7}{12}\right):\left(-\frac{38}{45}\right);$$
2. $$\left(\frac{11}{16}-\left(-\frac{17}{24}\right)\right):\left(-5\frac{7}{12}\right);$$
3. $$\left(-\frac{11}{18}+\frac{29}{45}\right):\left(\frac{35}{54}-\frac{19}{27}\right);$$
4. $$-4\frac{1}{7}+2\frac{1}{4}\cdot\left(-11\frac{2}{9}-\left(-5{,}4\right):\frac{9}{35}\right).$$

1)

$$\left(-\frac{4}{15}+\frac{7}{12}\right):\left(-\frac{38}{45}\right) = \left(-\frac{16}{60}+\frac{35}{60}\right):\left(-\frac{38}{45}\right) = \frac{19}{60}:\left(-\frac{38}{45}\right)$$
$$=\frac{19}{60}\cdot\left(-\frac{45}{38}\right) = -\frac{19\cdot 45}{60\cdot 38} = -\frac{3}{8}$$

2)

$$\left(\frac{11}{16}-\left(-\frac{17}{24}\right)\right):\left(-5\frac{7}{12}\right) = \left(\frac{33}{48}+\frac{34}{48}\right):\left(-\frac{67}{12}\right) = \frac{67}{48}:\left(-\frac{67}{12}\right)$$
$$= \frac{67}{48}\cdot\left(-\frac{12}{67}\right) = -\frac{1}{4}$$

3)

$$\left(-\frac{11}{18}+\frac{29}{45}\right):\left(\frac{35}{54}-\frac{19}{27}\right) = \left(-\frac{55}{90}+\frac{58}{90}\right):\left(\frac{35}{54}-\frac{38}{54}\right)$$
$$= \frac{3}{90}:\left(-\frac{3}{54}\right) = \frac{1}{30}:\left(-\frac{1}{18}\right) = \frac{1}{30}\cdot(-18) = -\frac{3}{5}$$

4)

$$-4\frac{1}{7}+2\frac{1}{4}\cdot\left(-11\frac{2}{9}-(-5{,}4):\frac{9}{35}\right)$$
$$=-4\frac{1}{7}+\frac{9}{4}\cdot\left(-11\frac{2}{9}+\frac{54}{10}:\frac{9}{35}\right)$$
$$=-4\frac{1}{7}+\frac{9}{4}\cdot\left(-11\frac{2}{9}+\frac{54}{10}\cdot\frac{35}{9}\right)$$
$$=-4\frac{1}{7}+\frac{9}{4}\cdot\left(-11\frac{2}{9}+21\right) =-4\frac{1}{7}+\frac{9}{4}\cdot 9\frac{7}{9}$$
$$=-4\frac{1}{7}+\frac{9}{4}\cdot\frac{88}{9} =-4\frac{1}{7}+22 =17\frac{6}{7}$$

Ответ: 1) $$-\frac{3}{8}$$; 2) $$-\frac{1}{4}$$; 3) $$-\frac{3}{5}$$; 4) $$17\frac{6}{7}$$.