Упр.199 Вариант 1 Дидактические материалы ГДЗ Мерзляк Полонский 6 класс (Математика)

Найдите значение выражения:

1. 1) $$\left(-\frac{7}{18}+\frac{11}{12}\right):\left(-\frac{19}{48}\right);$$
   2) $$\left(\frac{7}{16}-\left(-\frac{5}{24}\right)\right):\left(-1\frac{15}{16}\right);$$
   3) $$\left(\frac{10}{21}-\frac{25}{28}\right):\left(-\frac{11}{14}+\frac{24}{35}\right);$$
   4) $$-2\frac{2}{3}+2\frac{1}{3}\cdot\left(-15\frac{3}{7}-\left(-4{,}8\right):\frac{4}{15}\right).$$

1) $$\left(-\frac{7}{18}+\frac{11}{12}\right):\left(-\frac{19}{48}\right)=\left(-\frac{14}{36}+\frac{33}{36}\right):\left(-\frac{19}{48}\right)=\frac{19}{36}:\left(-\frac{19}{48}\right)$$

$$\frac{19}{36}\cdot\left(-\frac{48}{19}\right)=-\frac{48}{36}=-\frac{4}{3}=-1\frac{1}{3}$$

2) $$\left(\frac{7}{16}-\left(-\frac{5}{24}\right)\right):\left(-1\frac{15}{16}\right)=\left(\frac{21}{48}+\frac{10}{48}\right):\left(-\frac{31}{16}\right)=\frac{31}{48}:\left(-\frac{31}{16}\right)$$

$$\frac{31}{48}\cdot\left(-\frac{16}{31}\right)=-\frac{16}{48}=-\frac{1}{3}$$

3) $$\left(\frac{10}{21}-\frac{25}{28}\right):\left(-\frac{11}{14}+\frac{24}{35}\right)=\left(\frac{40}{84}-\frac{75}{84}\right):\left(-\frac{55}{70}+\frac{48}{70}\right)$$

$$-\frac{35}{84}:\left(-\frac{7}{70}\right)=\frac{35}{84}:\frac{7}{70}=\frac{35}{84}\cdot\frac{70}{7}=\frac{25}{6}=4\frac{1}{6}$$

4) $$-2\frac{2}{3}+2\frac{1}{3}\cdot\left(-15\frac{3}{7}-(-4{,}8):\frac{4}{15}\right)$$

$$-4{,}8=-\frac{24}{5},\qquad \left(-4{,}8\right):\frac{4}{15}=-\frac{24}{5}\cdot\frac{15}{4}=-18$$

Тогда

$$-15\frac{3}{7}-(-18)=-15\frac{3}{7}+18=2\frac{4}{7}=\frac{18}{7}$$

$$-2\frac{2}{3}+2\frac{1}{3}\cdot\frac{18}{7}=-\frac{8}{3}+\frac{7}{3}\cdot\frac{18}{7}=-\frac{8}{3}+6=\frac{10}{3}=3\frac{1}{3}$$

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