Упр.17.17 Часть 1 ГДЗ Мордкович Семенов 7 класс (Алгебра)

а) (1 3/7 — 1,4 · 2 6/7) · 0,7 + 1 3/5;
б) 2,7 — 3/4 : 5/13 + 1 7/12 · 10/19.

а)

$$\left(1\frac{3}{7}-1{,}4\cdot 2\frac{6}{7}\right)\cdot 0{,}7+1\frac{3}{5}$$
$$=\left(\frac{10}{7}-\frac{7}{5}\cdot\frac{20}{7}\right)\cdot\frac{7}{10}+\frac{8}{5}$$
$$=\left(\frac{10}{7}-4\right)\cdot\frac{7}{10}+\frac{8}{5}$$
$$=\left(\frac{10}{7}-\frac{28}{7}\right)\cdot\frac{7}{10}+\frac{8}{5}$$
$$=-\frac{18}{7}\cdot\frac{7}{10}+\frac{8}{5}=-\frac{18}{10}+\frac{8}{5}=-1{,}8+1{,}6=-0{,}2.$$

б)

$$2{,}7-\frac{3}{4}:\frac{5}{13}+1\frac{7}{12}\cdot\frac{10}{19}$$
$$=2{,}7-\frac{3}{4}\cdot\frac{13}{5}+\frac{19}{12}\cdot\frac{10}{19}$$
$$=2{,}7-\frac{39}{20}+\frac{5}{6}$$
$$=2\frac{7}{10}-1\frac{19}{20}+\frac{5}{6}$$
$$=\frac{162}{60}-\frac{117}{60}+\frac{50}{60}=\frac{95}{60}=\frac{19}{12}=1\frac{7}{12}.$$

Ответ

а) $$-0{,}2$$; б) $$1\frac{7}{12}$$.