Упр.1.2 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1. 1.2. Вычислите значение числового выражения:
   1) $$14\frac{7}{15}-3\frac{3}{23}\cdot\frac{23}{27}-1\frac{1}{5}\cdot\frac{1}{6}$$;
   2) $$\left(5\frac{8}{9}:1\frac{17}{36}+1\frac{1}{4}\right)\cdot\frac{5}{21}$$;
   3) $$(-3{,}25-2{,}75):(-0{,}6)+0{,}8\cdot(-7)$$;
   4) $$\left(-1\frac{3}{8}-2\frac{5}{12}\right):5\frac{5}{12}$$.

1. $$14\frac{7}{15}-3\frac{3}{23}\cdot\frac{23}{27}-1\frac{1}{5}\cdot\frac{1}{6}$$
   $$=14\frac{7}{15}-\frac{72}{23}\cdot\frac{23}{27}-\frac{6}{5}\cdot\frac{1}{6}$$
   $$=14\frac{7}{15}-\frac{8}{3}-\frac{1}{5}$$
   $$=14\frac{7}{15}-2\frac{10}{15}-\frac{3}{15}=11\frac{9}{15}=11\frac{3}{5}.$$
2. $$\left(5\frac{8}{9}:1\frac{17}{36}+1\frac{1}{4}\right)\cdot\frac{5}{21}$$
   $$=\left(\frac{53}{9}:\frac{53}{36}+1\frac{1}{4}\right)\cdot\frac{5}{21}$$
   $$=\left(\frac{53}{9}\cdot\frac{36}{53}+1\frac{1}{4}\right)\cdot\frac{5}{21}$$
   $$=\left(4+1\frac{1}{4}\right)\cdot\frac{5}{21}$$
   $$=5\frac{1}{4}\cdot\frac{5}{21}=\frac{21}{4}\cdot\frac{5}{21}=\frac{5}{4}=1\frac{1}{4}.$$
3. $$(-3{,}25-2{,}75):(-0{,}6)+0{,}8\cdot(-7)$$
   $$=-6:(-0{,}6)-5{,}6$$
   $$=10-5{,}6=4{,}4.$$
4. $$\left(-1\frac{3}{8}-2\frac{5}{12}\right):5\frac{5}{12}$$
   $$=\left(-\frac{11}{8}-\frac{29}{12}\right):\frac{65}{12}$$
   $$=\left(-\frac{33}{24}-\frac{58}{24}\right):\frac{65}{12}$$
   $$=-\frac{91}{24}:\frac{65}{12}=-\frac{91}{24}\cdot\frac{12}{65}=-\frac{7}{10}=-0{,}7.$$

Ответ

1) $$11\frac{3}{5}$$; 2) $$1\frac{1}{4}$$; 3) $$4{,}4$$; 4) $$-0{,}7$$.