Упр.184 Вариант 1 Дидактические материалы ГДЗ Мерзляк Полонский 6 класс (Математика)

1) -14,3 · 0,6 + 5,7 · (-1,4);
2) (3,4 — 5) · (-4,12 + 4,04);
3) 5/9 · (-3 6/7) — (-3 5/7) · 3/52;
4) (2 1/3 — 3 1/4) · (2 3/4 — 0,95).

1) $$-14{,}3 \cdot 0{,}6 + 5{,}7 \cdot (-1{,}4)$$

$$-14{,}3 \cdot 0{,}6 = -8{,}58$$

$$5{,}7 \cdot (-1{,}4) = -7{,}98$$

$$-8{,}58 + (-7{,}98) = -16{,}56$$

2) $$\left(3{,}4 — 5\right)\cdot\left(-4{,}12 + 4{,}04\right)$$

$$3{,}4 — 5 = -1{,}6$$

$$-4{,}12 + 4{,}04 = -0{,}08$$

$$(-1{,}6)\cdot(-0{,}08)=0{,}128$$

3) $$\frac{5}{9}\cdot\left(-3\frac{6}{7}\right)-\left(-3\frac{5}{7}\right)\cdot\frac{3}{52}$$

$$-3\frac{6}{7}=-\frac{27}{7}, \qquad -3\frac{5}{7}=-\frac{26}{7}$$

$$\frac{5}{9}\cdot\left(-\frac{27}{7}\right)-\left(-\frac{26}{7}\right)\cdot\frac{3}{52}$$

$$=-\frac{5\cdot 27}{9\cdot 7}+\frac{26\cdot 3}{7\cdot 52}$$

$$=-\frac{15}{7}+\frac{3}{14}=-2\frac{1}{7}+\frac{3}{14}=-1\frac{13}{14}$$

4) $$\left(2\frac{1}{3}-3\frac{1}{4}\right)\cdot\left(2\frac{3}{4}-0{,}95\right)$$

$$2\frac{1}{3}=\frac{7}{3}, \qquad 3\frac{1}{4}=\frac{13}{4}, \qquad 2\frac{3}{4}=\frac{11}{4}$$

$$\left(\frac{7}{3}-\frac{13}{4}\right)\cdot(2{,}75-0{,}95)$$

$$\frac{7}{3}-\frac{13}{4}=\frac{28-39}{12}=-\frac{11}{12}$$

$$2{,}75-0{,}95=1{,}8$$

$$-\frac{11}{12}\cdot 1{,}8=-\frac{11}{12}\cdot\frac{18}{10}=-\frac{11}{12}\cdot\frac{9}{5}=-\frac{33}{20}=-1\frac{13}{20}$$

Ответ

1) $$-16{,}56$$; 2) $$0{,}128$$; 3) $$-1\frac{13}{14}$$; 4) $$-1\frac{13}{20}$$.