Упр.755 Часть 3 ГДЗ Дорофеев Петерсон 6 класс (Математика)

Задача

а)-5 1/4•3/7 :(-1,8)•(-2/5)^2;
б) (-0,7•0,06•(-1,2))/(0,024•(-1/6)•5,6);
в) (-9/14 :(-5 1/7)+9,31:(-24,5)) :(-0,5)^2-0,08;
г) ((-2,75•2 2/11-1 7/8•(-3,2)) :(2 9/40-0,275))/(2,47:0,26-(-5 3/11 :5 3/11)^2 ).

Подробный ответ

а)

$$ -5\frac{1}{4}\cdot \frac{3}{7}:(-1{,}8)\cdot \left(-\frac{2}{5}\right)^2 = -\frac{21}{4}\cdot \frac{3}{7}:\left(-\frac{9}{5}\right)\cdot \frac{4}{25} $$
$$ = -\frac{63}{28}:\left(-\frac{9}{5}\right)\cdot \frac{4}{25} = -\frac{9}{4}\cdot \left(-\frac{5}{9}\right)\cdot \frac{4}{25} = \frac{1}{5}=0{,}2 $$

б)

$$ \frac{-0{,}7\cdot 0{,}06\cdot (-1{,}2)}{0{,}024\cdot \left(-\frac{1}{6}\right)\cdot 5{,}6} = \frac{\frac{7}{10}\cdot \frac{6}{100}\cdot \frac{12}{10}}{\frac{24}{1000}\cdot \left(-\frac{1}{6}\right)\cdot \frac{56}{10}} $$
$$ = \frac{7\cdot 6\cdot 12}{24\cdot (-1)\cdot 56} = \frac{7\cdot 3}{-1\cdot 8} = -\frac{21}{8} = -2{,}625 $$

в)

$$ \left(-\frac{9}{14}:(-5\frac{1}{7})+9{,}31:(-24{,}5)\right):(-0{,}5)^2-0{,}08 $$
$$ = \left(-\frac{9}{14}:\left(-\frac{36}{7}\right)+\frac{931}{100}:\left(-\frac{49}{2}\right)\right):\frac{1}{4}-0{,}08 $$
$$ = \left(\frac{9}{14}\cdot \frac{7}{36}-\frac{931}{100}\cdot \frac{2}{49}\right):\frac{1}{4}-0{,}08 = \left(\frac{1}{8}-0{,}38\right)\cdot 4-0{,}08 $$
$$ = (-0{,}255)\cdot 4-0{,}08 = -1{,}02-0{,}08 = -1{,}1 $$

г)

$$ \frac{\left(-2{,}75\cdot 2\frac{2}{11}-1\frac{7}{8}\cdot (-3{,}2)\right):\left(2\frac{9}{40}-0{,}275\right)}{2{,}47:0{,}26-\left(-5\frac{3}{11}:5\frac{3}{11}\right)^2} $$
$$ = \frac{\left(-\frac{11}{4}\cdot \frac{24}{11}+\frac{15}{8}\cdot \frac{32}{10}\right):\left(2\frac{9}{40}-0{,}275\right)}{9{,}5-(-1)^2} $$
$$ = \frac{(-6+6):1{,}95}{9{,}5-1} = \frac{0}{8{,}5} = 0 $$