Упр.1 Повторение ГДЗ Никольский Потапов 11 класс (Алгебра)

1. (МГУЭСИ).

1. $$\frac{3+4{,}2:0{,}1}{(1:0{,}3-2\frac{1}{3})\cdot 0{,}3125}$$

   $$4{,}2:0{,}1=42,\quad 1:0{,}3=\frac{10}{3},\quad 2\frac{1}{3}=\frac{7}{3}$$

   $$ \frac{3+42}{\left(\frac{10}{3}-\frac{7}{3}\right)\cdot 0{,}3125} = \frac{45}{1\cdot 0{,}3125} = 144 $$

2. $$\frac{0{,}134+0{,}05}{18\frac{1}{6}-1\frac{11}{14}-2\frac{6}{7}\cdot \frac{2}{15}}$$

   $$0{,}134+0{,}05=0{,}184$$

   $$ 2\frac{6}{7}\cdot \frac{2}{15} = \frac{20}{7}\cdot \frac{2}{15} = \frac{8}{21} $$

   $$ 18\frac{1}{6}-1\frac{11}{14}-\frac{8}{21} = \frac{109}{6}-\frac{25}{14}-\frac{8}{21} = \frac{1526-150-32}{84} = \frac{1344}{84} = 16 $$

   $$ \frac{0{,}184}{16}=0{,}0115 $$

3. $$417\cdot \left(\frac{2}{10}+\frac{13}{990}\right):\left(\frac{4}{10}+\frac{21}{990}\right)$$

   $$ \frac{2}{10}+\frac{13}{990} = \frac{198+13}{990} = \frac{211}{990} $$

   $$ \frac{4}{10}+\frac{21}{990} = \frac{396+21}{990} = \frac{417}{990} $$

   $$ 417\cdot \frac{211}{990}:\frac{417}{990} = 211 $$

4. $$78\cdot \frac{\left(4\frac{3}{5}-1\frac{3}{14}\right)\cdot 5\frac{5}{6}}{(11-1{,}25):2{,}5}$$

   $$ 4\frac{3}{5}-1\frac{3}{14} = 3\frac{3}{5}-\frac{3}{14} = 3+\frac{42-15}{70} = 3\frac{27}{70} $$

   $$ 3\frac{27}{70}-5\frac{5}{6} = \frac{237}{70}-\frac{35}{6} = \frac{79}{2}\cdot \frac{1}{2} = \frac{79}{4} $$

   $$ (11-1{,}25):2{,}5=9{,}75:2{,}5=3{,}9 $$

   $$ 78\cdot \frac{79}{4}:3{,}9 = 20\cdot \frac{79}{4} = 79\cdot 5 = 395 $$

Ответ

а) 144; б) 0,0115; в) 211; г) 395.