Упр.2 Задачи на повторение ГДЗ Мордкович 8 класс (Алгебра)

a) 1/2 + 2 2/3 + 1 1/2 + 1 1/3; в) (3/14 — 2/7 + 1/2) · 14;
б) 3 2/5 · 2 3/7 · 5 · 7; г) (12 2/9 + 24 2/3 — 16 2/15) : 2.

1. $$\frac12+2\frac23+1\frac12+1\frac13=\left(\frac12+1\frac12\right)+\left(2\frac23+1\frac13\right)=2+4=6.$$

2. $$3\frac25\cdot 2\frac37\cdot 5\cdot 7=\left(3\frac25\cdot 5\right)\left(2\frac37\cdot 7\right)=\left(\frac{17}{5}\cdot 5\right)\left(\frac{17}{7}\cdot 7\right)=17\cdot 17=289.$$

3. $$\left(\frac{3}{14}-\frac27+\frac12\right)\cdot 14=\frac{3}{14}\cdot 14-\frac27\cdot 14+\frac12\cdot 14=3-4+7=6.$$

4. $$\left(12\frac29+24\frac23-16\frac{2}{15}\right):2=\left(12\frac{10}{45}+24\frac{30}{45}-16\frac{6}{45}\right)\cdot\frac12$$

   $$=\left(36\frac{40}{45}-16\frac{6}{45}\right)\cdot\frac12=20\frac{34}{45}\cdot\frac12=\left(20+\frac{34}{45}\right)\cdot\frac12$$

   $$=20\cdot\frac12+\frac{34}{45}\cdot\frac12=10+\frac{17}{45}=10\frac{17}{45}.$$

Ответ

а) $$6$$; б) $$289$$; в) $$6$$; г) $$10\frac{17}{45}$$.