Упр.10 Повторение ГДЗ Мордкович Семенов 7 класс (Алгебра)

Задача

Найдите значение выражения:

а) $$\left(0{,}56 : 1{,}4 — 2\frac{2}{3} : 2\frac{2}{15}\right)\cdot\left(-\frac{6}{17}\right)$$;

б) $$\left(-4{,}1 + 5{,}67 — 9{,}57\right)\cdot\left(-\frac{1}{2}\right)^4$$;

в) $$\left(-0{,}72 : 2{,}4 + 8\frac{4}{7}\cdot 4{,}2\right) : \left(-1\frac{4}{7}\right)$$;

г) $$\left(-\frac{1}{4}\right)^3 : \left(-1{,}3 + 3{,}8 — 3\frac{1}{8}\right)$$.

Подробный ответ

а)

$$(0{,}56 : 1{,}4 — 2\tfrac{2}{3} : 2\tfrac{2}{15})\cdot\left(-\tfrac{6}{17}\right)$$
$$0{,}56 : 1{,}4 = 0{,}4,\qquad 2\tfrac{2}{3}=\tfrac{8}{3},\qquad 2\tfrac{2}{15}=\tfrac{32}{15}$$
$$\tfrac{8}{3} : \tfrac{32}{15}=\tfrac{8}{3}\cdot\tfrac{15}{32}=\tfrac{5}{4}=1{,}25$$
$$0{,}4-1{,}25=-0{,}85=-\tfrac{17}{20}$$
$$-\tfrac{17}{20}\cdot\left(-\tfrac{6}{17}\right)=\tfrac{6}{20}=\tfrac{3}{10}=0{,}3$$

б)

$$(-4{,}1+5{,}67-9{,}57)\cdot\left(-\tfrac{1}{2}\right)^4$$
$$-4{,}1+5{,}67-9{,}57=-8$$
$$\left(-\tfrac{1}{2}\right)^4=\tfrac{1}{16}$$
$$-8\cdot\tfrac{1}{16}=-\tfrac{1}{2}=-0{,}5$$

в)

$$(-0{,}72:2{,}4+8\tfrac{4}{7}\cdot4{,}2):\left(-1\tfrac{4}{7}\right)$$
$$-0{,}72:2{,}4=-0{,}3,\qquad 8\tfrac{4}{7}=\tfrac{60}{7},\qquad 4{,}2=\tfrac{21}{5}$$
$$\tfrac{60}{7}\cdot4{,}2=\tfrac{60}{7}\cdot\tfrac{21}{5}=36$$
$$-0{,}3+36=35{,}7$$
$$-1\tfrac{4}{7}=-\tfrac{11}{7}$$
$$35{,}7:\left(-\tfrac{11}{7}\right)=\tfrac{357}{10}\cdot\left(-\tfrac{7}{11}\right)=-\tfrac{2499}{110}=-22\tfrac{79}{110}$$

г)

$$\left(-\tfrac{1}{4}\right)^3:\left(-1{,}3+3{,}8-3\tfrac{1}{8}\right)$$
$$\left(-\tfrac{1}{4}\right)^3=-\tfrac{1}{64}$$
$$-1{,}3+3{,}8-3\tfrac{1}{8}=2{,}5-3{,}125=-0{,}625=-\tfrac{5}{8}$$
$$-\tfrac{1}{64}:\left(-\tfrac{5}{8}\right)=-\tfrac{1}{64}\cdot\left(-\tfrac{8}{5}\right)=\tfrac{1}{40}$$