Упр.1 Повторение(старый учебник) ГДЗ Мордкович Александрова 9 класс (Алгебра)
а) (8*7/12 — 2*17/36) *2,7 — 4*1/3 :0,65;
б) (1*11/24 + 13/36) * 1,44 — 8/15*0,5625.
а)
$$\left(8\frac{7}{12}-2\frac{17}{36}\right)\cdot 2{,}7-4\frac{1}{3}:0{,}65$$
$$8\frac{7}{12}=\frac{103}{12}, \qquad 2\frac{17}{36}=\frac{89}{36}, \qquad 2{,}7=\frac{27}{10}, \qquad 0{,}65=\frac{13}{20}$$
$$\left(\frac{103}{12}-\frac{89}{36}\right)\cdot \frac{27}{10}-4\frac{1}{3}:\frac{13}{20}$$
$$\frac{309-89}{36}\cdot \frac{27}{10}-\frac{13}{3}\cdot \frac{20}{13}$$
$$\frac{220}{36}\cdot \frac{27}{10}-\frac{20}{3}=\frac{55}{9}\cdot \frac{27}{10}-\frac{20}{3}=\frac{33}{2}-\frac{20}{3}=\frac{99-40}{6}=\frac{59}{6}=9\frac{5}{6}$$
б)
$$\left(1\frac{11}{24}+\frac{13}{36}\right)\cdot 1{,}44-\frac{8}{15}\cdot 0{,}5625$$
$$1\frac{11}{24}=\frac{35}{24}, \qquad 1{,}44=\frac{144}{100}, \qquad 0{,}5625=\frac{9}{16}$$
$$\left(\frac{35}{24}+\frac{13}{36}\right)\cdot \frac{144}{100}-\frac{8}{15}\cdot \frac{9}{16}$$
$$\frac{105+26}{72}\cdot \frac{144}{100}-\frac{72}{240}=\frac{131}{72}\cdot \frac{144}{100}-\frac{3}{10}$$
$$\frac{131\cdot 2}{100}-\frac{3}{10}=\frac{262}{100}-\frac{30}{100}=\frac{232}{100}=2{,}32$$
Ответ
а) $$9\frac{5}{6}$$; б) $$2{,}32$$.
