Упр.686 ГДЗ Зубарева Мордкович 6 класс ФГОС (Математика)

Задача

а) (13/17•8/9-4/17•8/9)•(11/16+3/8);
б) (3/4-15/16)•(1-3 2/7);
в) (1 5/12-2 1/8) :(1 1/2+23/30);
г) (2 2/15-3 3/5) :(14/15+2/45).

Подробный ответ

а)

$$ \left(\frac{13}{17}\cdot\frac{8}{9}-\frac{4}{17}\cdot\frac{8}{9}\right)\cdot\left(\frac{11}{16}+\frac{3}{8}\right) $$
$$ =\left(\frac{8}{9}\cdot\left(\frac{13}{17}-\frac{4}{17}\right)\right)\cdot\left(\frac{11}{16}+\frac{6}{16}\right) $$
$$ =\left(\frac{8}{9}\cdot\frac{9}{17}\right)\cdot\frac{17}{16} =\frac{8}{17}\cdot\frac{17}{16} =\frac{8}{16} =\frac{1}{2} $$

б)

$$ \left(\frac{3}{4}-\frac{15}{16}\right)\cdot\left(1-3\frac{2}{7}\right) $$
$$ =\left(\frac{12}{16}-\frac{15}{16}\right)\cdot\left(\frac{7}{7}-\frac{23}{7}\right) $$
$$ =-\frac{3}{16}\cdot\left(-\frac{16}{7}\right) =\frac{3}{7} $$

в)

$$ \left(1\frac{5}{12}-2\frac{1}{8}\right):\left(1\frac{1}{2}+\frac{23}{30}\right) $$
$$ =\left(\frac{17}{12}-\frac{17}{8}\right):\left(\frac{3}{2}+\frac{23}{30}\right) $$
$$ =\left(\frac{34}{24}-\frac{51}{24}\right):\left(\frac{45}{30}+\frac{23}{30}\right) $$
$$ =-\frac{17}{24}:\frac{68}{30} =-\frac{17}{24}\cdot\frac{30}{68} =-\frac{5}{16} $$

г)

$$ \left(2\frac{2}{15}-3\frac{3}{5}\right):\left(\frac{14}{15}+\frac{2}{45}\right) $$
$$ =\left(\frac{32}{15}-\frac{18}{5}\right):\left(\frac{42}{45}+\frac{2}{45}\right) $$
$$ =\left(\frac{32}{15}-\frac{54}{15}\right):\frac{44}{45} $$
$$ =-\frac{22}{15}:\frac{44}{45} =-\frac{22}{15}\cdot\frac{45}{44} =-\frac{3}{2} =-1{,}5 $$