Упр.977 ГДЗ Зубарева Мордкович 6 класс ФГОС (Математика)

Задача

а) (1/3+0,5)•2 2/5-(1/2-0,3)•3,5;
б) 7:(1 1/5-2,2)+1/10 :(2,8-3/8);
в) (1-7/15+0,4) :(8,3+4/19-15/38);
г) (4,25•16/51-17/18 :8,5)•(3/11)^2.

Подробный ответ

а) Преобразуем десятичные дроби в обыкновенные:

$$ \left(\frac13+0{,}5\right)\cdot 2\frac25-\left(\frac12-0{,}3\right)\cdot 3{,}5 = \left(\frac13+\frac12\right)\cdot \frac{12}{5}-\left(\frac12-\frac{3}{10}\right)\cdot \frac{7}{2} $$
$$ = \frac56\cdot \frac{12}{5}-\frac15\cdot \frac72 = 2-\frac{7}{10} = 1{,}3 $$

б)

$$ 7:\left(1\frac15-2{,}2\right)+\frac{1}{10}:(2{,}8-\frac38) = 7:(1{,}2-2{,}2)+\frac{1}{10}:\left(2\frac45-\frac38\right) $$
$$ = 7:(-1)+\frac{1}{10}:\left(\frac{14}{5}-\frac38\right) = -7+\frac{1}{10}:\frac{97}{40} $$
$$ = -7+\frac{1}{10}\cdot \frac{40}{97} = -7+\frac{4}{97} = -\frac{679}{97} = -6\frac{97}{97} = -6\frac{93}{97} $$

в)

$$ \left(1-\frac{7}{15}+0{,}4\right):\left(8{,}3+\frac{4}{19}-\frac{15}{38}\right) = \left(1-\frac{7}{15}+\frac25\right):\left(8{,}3+\frac{8}{38}-\frac{15}{38}\right) $$
$$ = \left(\frac{8}{15}+\frac25\right):\left(8\frac{3}{10}-\frac{7}{38}\right) = \frac{14}{15}:\left(8\frac{57}{190}-\frac{35}{190}\right) $$
$$ = \frac{14}{15}:\,8\frac{22}{190} = \frac{14}{15}:\frac{771}{95} = \frac{14}{15}\cdot \frac{95}{771} = \frac{14\cdot 19}{3\cdot 771} = \frac{266}{2313} $$

г)

$$ \left(4{,}25\cdot \frac{16}{51}-\frac{17}{18}:8{,}5\right)\cdot \left(\frac{3}{11}\right)^2 = \left(\frac{17}{4}\cdot \frac{16}{51}-\frac{17}{18}:\frac{17}{2}\right)\cdot \frac{9}{121} $$
$$ = \left(\frac{4}{3}-\frac{1}{9}\right)\cdot \frac{9}{121} = \frac{11}{9}\cdot \frac{9}{121} = \frac{1}{11} $$