Упр.19.11 ГДЗ Рабочая тетрадь Зубарева Мордкович 6 класс Экзамен (Математика)

Задача

а) 1/7•(-3/5)+1/7•(-2/5);
б) 0,84•6+0,16•6;
в) 0,77•23+0,77•77;
г) 3/7•4/17-3/7•4/17;
д) 16/27•(-25/72)+16/27•(-11/72);
е) 19/20•(2/19-3/7)+19/20•3/7;
ж) 11/13•(24/37-13/11)+24/37•(-11/13+37/24);
з) 3/5•(6/7-5/3)+6/5•(5/6-3/7);
и) 11/13•(-3/7+13/22)+3/13•(-11/7-13/3).

Подробный ответ

Используем распределительное свойство умножения и группируем слагаемые с общим множителем.

а)
$$ \frac{1}{7}\cdot\left(-\frac{3}{5}\right)+\frac{1}{7}\cdot\left(-\frac{2}{5}\right) =\frac{1}{7}\cdot\left(-\frac{3}{5}-\frac{2}{5}\right) =\frac{1}{7}\cdot(-1) =-\frac{1}{7} $$

б)
$$ 0{,}84\cdot 6+0{,}16\cdot 6 =6\cdot(0{,}84+0{,}16) =6\cdot 1 =6 $$

в)
$$ 0{,}77\cdot 23+0{,}77\cdot 77 =0{,}77\cdot(23+77) =0{,}77\cdot 100 =77 $$

г)
$$ \frac{3}{7}\cdot\frac{4}{17}-\frac{3}{7}\cdot\frac{4}{17}=0 $$

д)
$$ \frac{16}{27}\cdot\left(-\frac{25}{72}\right)+\frac{16}{27}\cdot\left(-\frac{11}{72}\right) =\frac{16}{27}\cdot\left(-\frac{25}{72}-\frac{11}{72}\right) =\frac{16}{27}\cdot\left(-\frac{36}{72}\right) =\frac{16}{27}\cdot\left(-\frac{1}{2}\right) =-\frac{8}{27} $$

е)
$$ \frac{19}{20}\cdot\left(\frac{2}{19}-\frac{3}{7}\right)+\frac{19}{20}\cdot\frac{3}{7} =\frac{19}{20}\cdot\left(\frac{2}{19}-\frac{3}{7}+\frac{3}{7}\right) =\frac{19}{20}\cdot\frac{2}{19} =\frac{1}{10} $$

ж)
$$ \frac{11}{13}\cdot\left(\frac{24}{37}-\frac{13}{11}\right)+\frac{24}{37}\cdot\left(-\frac{11}{13}+\frac{37}{24}\right) $$
$$ =\frac{11}{13}\cdot\frac{24}{37}-\frac{11}{13}\cdot\frac{13}{11}-\frac{24}{37}\cdot\frac{11}{13}+\frac{24}{37}\cdot\frac{37}{24} $$
$$ =\frac{11}{13}\cdot\frac{24}{37}-1-\frac{24}{37}\cdot\frac{11}{13}+1=0 $$

з)
$$ \frac{3}{5}\cdot\left(\frac{6}{7}-\frac{5}{3}\right)+\frac{6}{5}\cdot\left(\frac{5}{6}-\frac{3}{7}\right) $$
$$ =\frac{3}{5}\cdot\frac{6}{7}-\frac{3}{5}\cdot\frac{5}{3}+\frac{6}{5}\cdot\frac{5}{6}-\frac{6}{5}\cdot\frac{3}{7} $$
$$ =\frac{3}{5}\cdot\frac{6}{7}-1+1-\frac{6}{5}\cdot\frac{3}{7}=0 $$

и)
$$ \frac{11}{13}\cdot\left(-\frac{3}{7}+\frac{13}{22}\right)+\frac{3}{13}\cdot\left(-\frac{11}{7}-\frac{13}{3}\right) $$
$$ =\frac{11}{13}\cdot\left(-\frac{3}{7}\right)+\frac{11}{13}\cdot\frac{13}{22}+\frac{3}{13}\cdot\left(-\frac{11}{7}\right)-\frac{3}{13}\cdot\frac{13}{3} $$
$$ =-\frac{33}{91}+\frac{1}{2}-\frac{33}{91}-1 =-\frac{66}{91}-\frac{1}{2} =-\frac{132}{182}-\frac{91}{182} =-\frac{223}{182} =-1\frac{41}{182} $$