Упр.820 ГДЗ Никольский Потапов 7 класс (Алгебра)

б) (4-1*1/6+6*1/4):1/2;
в) 3*3/5 — 1*1/2+2*2/5;
г) 7*3/8-4*3/4+3*1/2;
д) (4*1/3 — 12*1/2-2*1/5):2;
е) (6*2/3-9*3/5+15*1/2):3.

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

   $$8\frac{1}{2}=8\frac{2}{4},\quad 8\frac{2}{4}-3\frac{3}{4}=4\frac{3}{4}$$

   $$4\frac{3}{4}\cdot 8=\frac{19}{4}\cdot 8=19\cdot 2=38$$

2. б) $$\left(4-1\frac{1}{6}+6\frac{1}{4}\right):\frac{1}{2}$$

   $$4=3\frac{12}{12},\quad 1\frac{1}{6}=1\frac{2}{12},\quad 6\frac{1}{4}=6\frac{3}{12}$$

   $$\left(3\frac{12}{12}-1\frac{2}{12}+6\frac{3}{12}\right):\frac{1}{2}=8\frac{13}{12}:\frac{1}{2}=\frac{109}{12}\cdot 2=\frac{109}{6}=18\frac{1}{6}$$

3. в) $$3\frac{3}{5}-1\frac{1}{2}+2\frac{2}{5}$$

   $$3\frac{3}{5}=3\frac{6}{10},\quad 1\frac{1}{2}=1\frac{5}{10},\quad 2\frac{2}{5}=2\frac{4}{10}$$

   $$3\frac{6}{10}-1\frac{5}{10}+2\frac{4}{10}=4\frac{5}{10}=4\frac{1}{2}$$

4. г) $$7\frac{3}{8}-4\frac{3}{4}+3\frac{1}{2}$$

   $$4\frac{3}{4}=4\frac{6}{8},\quad 3\frac{1}{2}=3\frac{4}{8}$$

   $$7\frac{3}{8}-4\frac{6}{8}+3\frac{4}{8}=6\frac{1}{8}$$

5. д) $$\left(4\frac{1}{3}-12\frac{1}{2}-2\frac{1}{5}\right):2$$

   $$4\frac{1}{3}=4\frac{10}{30},\quad 12\frac{1}{2}=12\frac{15}{30},\quad 2\frac{1}{5}=2\frac{6}{30}$$

   $$\left(4\frac{10}{30}-12\frac{15}{30}-2\frac{6}{30}\right):2=-10\frac{11}{30}:2=-\frac{311}{30}\cdot\frac{1}{2}=-\frac{311}{60}=-5\frac{11}{60}$$

6. е) $$\left(6\frac{2}{3}-9\frac{3}{5}+15\frac{1}{2}\right):3$$

   $$6\frac{2}{3}=6\frac{20}{30},\quad 9\frac{3}{5}=9\frac{18}{30},\quad 15\frac{1}{2}=15\frac{15}{30}$$

   $$\left(6\frac{20}{30}-9\frac{18}{30}+15\frac{15}{30}\right):3=12\frac{17}{30}:3=\frac{377}{30}\cdot\frac{1}{3}=\frac{377}{90}=4\frac{17}{90}$$

Ответ

а) $$38$$; б) $$18\frac{1}{6}$$; в) $$4\frac{1}{2}$$; г) $$6\frac{1}{8}$$; д) $$-5\frac{11}{60}$$; е) $$4\frac{17}{90}$$.