Упр.1.42 ГДЗ Мордкович 7 класс (Алгебра)

Найдите значение числового выражения:

а)

$$\left(8\frac{7}{12}-2\frac{17}{36}\right)\cdot 2{,}7-4\frac{1}{3}:0{,}65$$
$$8\frac{7}{12}-2\frac{17}{36}=6\frac{1}{9}$$
$$6\frac{1}{9}\cdot 2{,}7=16\frac{1}{2}$$
$$4\frac{1}{3}:0{,}65=6\frac{2}{3}$$
$$16\frac{1}{2}-6\frac{2}{3}=9\frac{5}{6}$$

б)

$$\left(1\frac{11}{24}+\frac{13}{36}\right)\cdot 1{,}44-\frac{8}{15}\cdot 0{,}5625$$
$$1\frac{11}{24}+\frac{13}{36}=1\frac{59}{72}$$
$$1\frac{59}{72}\cdot 1{,}44=2\frac{31}{50}$$
$$\frac{8}{15}\cdot 0{,}5625=\frac{3}{10}$$
$$2\frac{31}{50}-\frac{3}{10}=2\frac{8}{25}$$

в)

$$\left(6\frac{8}{15}-4\frac{21}{45}\right)\cdot 4{,}5-2\frac{1}{6}:0{,}52$$
$$6\frac{8}{15}-4\frac{21}{45}=2\frac{1}{15}$$
$$2\frac{1}{15}\cdot 4{,}5=9\frac{3}{10}$$
$$2\frac{1}{6}:0{,}52=4\frac{1}{6}$$
$$9\frac{3}{10}-4\frac{1}{6}=5\frac{2}{15}$$

г)

$$\left(\frac{9}{22}+1\frac{12}{33}\right)\cdot 1{,}32-\frac{8}{13}\cdot 0{,}1625$$
$$\frac{9}{22}+1\frac{12}{33}=1\frac{51}{66}$$
$$1\frac{51}{66}\cdot 1{,}32=2\frac{17}{50}$$
$$\frac{8}{13}\cdot 0{,}1625=\frac{1}{10}$$
$$2\frac{17}{50}-\frac{1}{10}=2\frac{6}{25}$$

Ответ

а) $$9\frac{5}{6}$$; б) $$2\frac{8}{25}$$; в) $$5\frac{2}{15}$$; г) $$2\frac{6}{25}$$.