Упр.1.7 ГДЗ Мордкович 7 класс (Алгебра)

б) (6-7*1/8)*(2/9+2/3);
в) 17:(4*1/3 — 3*1/5);
г) (15-4*1/8)*(3*14/15 — 2*3/5).

1. $$\left(4\frac{1}{3}+3\frac{1}{5}\right):113$$

   $$4\frac{1}{3}=\frac{13}{3},\quad 3\frac{1}{5}=\frac{16}{5}$$

   $$\frac{13}{3}+\frac{16}{5}=\frac{65+48}{15}=\frac{113}{15}$$

   $$\frac{113}{15}:113=\frac{113}{15}\cdot\frac{1}{113}=\frac{1}{15}$$

2. $$\left(6-7\frac{1}{8}\right)\cdot\left(\frac{2}{9}+\frac{2}{3}\right)$$

   $$6-7\frac{1}{8}=6-\frac{57}{8}=\frac{48-57}{8}=-\frac{9}{8}$$

   $$\frac{2}{9}+\frac{2}{3}=\frac{2}{9}+\frac{6}{9}=\frac{8}{9}$$

   $$-\frac{9}{8}\cdot\frac{8}{9}=-1$$

3. $$17:\left(4\frac{1}{3}-3\frac{1}{5}\right)$$

   $$4\frac{1}{3}=\frac{13}{3},\quad 3\frac{1}{5}=\frac{16}{5}$$

   $$\frac{13}{3}-\frac{16}{5}=\frac{65-48}{15}=\frac{17}{15}$$

   $$17:\frac{17}{15}=17\cdot\frac{15}{17}=15$$

4. $$\left(15-4\frac{1}{8}\right)\cdot\left(3\frac{14}{15}-2\frac{3}{5}\right)$$

   $$15-4\frac{1}{8}=15-\frac{33}{8}=\frac{120-33}{8}=\frac{87}{8}$$

   $$3\frac{14}{15}-2\frac{3}{5}=\frac{59}{15}-\frac{13}{5}=\frac{59}{15}-\frac{39}{15}=\frac{20}{15}=\frac{4}{3}$$

   $$\frac{87}{8}\cdot\frac{4}{3}=\frac{29}{2}=14\frac{1}{2}$$

Ответ

а) $$\frac{1}{15}$$; б) $$-1$$; в) $$15$$; г) $$14\frac{1}{2}$$.