Упр.19.11 ГДЗ Рабочая тетрадь Зубарева Мордкович 6 класс Экзамен (Математика)
- Используя распределительное свойство умножения, найдите значение выражения:
а) $$\frac{1}{7}\cdot\left(-\frac{3}{5}\right)+\frac{1}{7}\cdot\left(-\frac{2}{5}\right)$$;
б) $$0{,}84\cdot 6+0{,}16\cdot 6$$;
в) $$0{,}77\cdot 23+0{,}77\cdot 77$$;
г) $$\frac{3}{7}\cdot\frac{4}{17}-\frac{3}{7}\cdot\frac{4}{17}$$;
д) $$\frac{16}{27}\cdot\left(-\frac{25}{72}\right)+\frac{16}{27}\cdot\left(-\frac{11}{72}\right)$$;
е) $$\frac{19}{20}\cdot\left(\frac{2}{19}-\frac{3}{7}\right)+\frac{19}{20}\cdot\frac{3}{7}$$;
ж) $$\frac{11}{13}\cdot\left(\frac{24}{37}-\frac{13}{11}\right)+\frac{24}{37}\cdot\left(-\frac{11}{13}+\frac{37}{24}\right)$$;
з) $$\frac{3}{5}\cdot\left(\frac{6}{7}-\frac{5}{3}\right)+\frac{6}{5}\cdot\left(\frac{5}{6}-\frac{3}{7}\right)$$;
и) $$\frac{11}{13}\cdot\left(-\frac{3}{7}+\frac{13}{22}\right)+\frac{3}{13}\cdot\left(-\frac{11}{7}-\frac{13}{3}\right)$$.
Используем распределительное свойство умножения:
$$\frac17\cdot\left(-\frac35\right)+\frac17\cdot\left(-\frac25\right) =\frac17\cdot\left(-\frac35-\frac25\right) =\frac17\cdot(-1) =-\frac17$$
$$0{,}84\cdot 6+0{,}16\cdot 6 =6\cdot(0{,}84+0{,}16) =6\cdot 1 =6$$
$$0{,}77\cdot 23+0{,}77\cdot 77 =0{,}77\cdot(23+77) =0{,}77\cdot 100 =77$$
$$\frac37\cdot\frac4{17}-\frac37\cdot\frac4{17}=0$$
$$\frac{16}{27}\cdot\left(-\frac{25}{72}\right)+\frac{16}{27}\cdot\left(-\frac{11}{72}\right) =\frac{16}{27}\cdot\left(-\frac{25}{72}-\frac{11}{72}\right) =\frac{16}{27}\cdot\left(-\frac{36}{72}\right) =\frac{16}{27}\cdot\left(-\frac12\right) =-\frac{8}{27}$$
$$\frac{19}{20}\cdot\left(\frac2{19}-\frac37\right)+\frac{19}{20}\cdot\frac37 =\frac{19}{20}\cdot\left(\frac2{19}-\frac37+\frac37\right) =\frac{19}{20}\cdot\frac2{19} =\frac1{10} =0{,}1$$
$$\frac{11}{13}\cdot\left(\frac{24}{37}-\frac{13}{11}\right)+\frac{24}{37}\cdot\left(-\frac{11}{13}+\frac{37}{24}\right) =\frac{11}{13}\cdot\frac{24}{37}-1-\frac{24}{37}\cdot\frac{11}{13}+1 =0$$
$$\frac35\cdot\left(\frac67-\frac53\right)+\frac65\cdot\left(\frac56-\frac37\right) =\frac35\cdot\frac67-\frac35\cdot\frac53+\frac65\cdot\frac56-\frac65\cdot\frac37 =0$$
$$\frac{11}{13}\cdot\left(-\frac37+\frac{13}{22}\right)+\frac3{13}\cdot\left(-\frac{11}{7}-\frac{13}{3}\right) =\frac{11}{13}\cdot\left(-\frac37\right)+\frac{11}{13}\cdot\frac{13}{22}+\frac3{13}\cdot\left(-\frac{11}{7}\right)-1$$
$$=-\frac{33}{91}+\frac12-\frac{33}{91}-1 =-\frac{66}{91}-\frac12 =-\frac{132}{182}-\frac{91}{182} =-\frac{223}{182} =-1\frac{41}{182}$$
Ответ: а) $$-\frac17$$; б) $$6$$; в) $$77$$; г) $$0$$; д) $$-\frac{8}{27}$$; е) $$\frac1{10}$$; ж) $$0$$; з) $$0$$; и) $$-1\frac{41}{182}$$.
