Упр.878 ГДЗ Ткачёва 6 класс (Математика)

Задача

1) 0,0075 · 10^2 — 3,2 · (0,08 — 0,2^2) + 40 : 10^3;
2) 16,4 : 10^3 + 4,1 · (3,25 — 2,12) — 0,86 · 5^2;
3) (-1 2/7)^2 · 7/27 — 0,6 : 1 1/2 — (-1 1/3)^3;
4) (7 — (-2 1/2)^2) : 1,35 — (-3/5)^3 · 10^3.

Подробный ответ

1) $$0{,}0075 \cdot 10^2 — 3{,}2 \cdot (0{,}08 — 0{,}2^2) + 40 : 10^3$$

$$0{,}2^2=0{,}04$$

$$0{,}08-0{,}04=0{,}04$$

$$0{,}0075 \cdot 10^2=0{,}0075 \cdot 100=0{,}75$$

$$40:10^3=40:1000=0{,}04$$

$$3{,}2 \cdot 0{,}04=0{,}128$$

$$0{,}75-0{,}128+0{,}04=0{,}662$$

2) $$16{,}4 : 10^3 + 4{,}1 \cdot (3{,}25 — 2{,}1^2) — 0{,}86 \cdot 5^2$$

$$2{,}1^2=4{,}41$$

$$3{,}25-4{,}41=-1{,}16$$

$$16{,}4:10^3=16{,}4:1000=0{,}0164$$

$$4{,}1 \cdot (-1{,}16)=-4{,}756$$

$$0{,}86 \cdot 5^2=0{,}86 \cdot 25=21{,}5$$

$$0{,}0164-4{,}756-21{,}5=-26{,}2396$$

3) $$\left(-1 \dfrac{2}{7}\right)^2 \cdot \dfrac{7}{27} — 0{,}6 : 1 \dfrac{1}{2} — \left(-1 \dfrac{1}{3}\right)^3$$

$$-1 \dfrac{2}{7}=-\dfrac{9}{7}, \quad \left(-\dfrac{9}{7}\right)^2=\dfrac{81}{49}$$

$$\dfrac{81}{49}\cdot \dfrac{7}{27}=\dfrac{3}{7}$$

$$0{,}6:1 \dfrac{1}{2}=\dfrac{3}{5}:\dfrac{3}{2}=\dfrac{3}{5}\cdot \dfrac{2}{3}=\dfrac{2}{5}$$

$$-1 \dfrac{1}{3}=-\dfrac{4}{3}, \quad \left(-\dfrac{4}{3}\right)^3=-\dfrac{64}{27}$$

$$\dfrac{3}{7}-\dfrac{2}{5}-\left(-\dfrac{64}{27}\right)=\dfrac{3}{7}-\dfrac{2}{5}+\dfrac{64}{27}$$

$$\dfrac{3}{7}-\dfrac{2}{5}=\dfrac{15-14}{35}=\dfrac{1}{35}$$

$$\dfrac{1}{35}+\dfrac{64}{27}=\dfrac{27}{945}+\dfrac{2240}{945}=\dfrac{2267}{945}=2 \dfrac{377}{945}$$

4) $$\bigl(7-(-2 \dfrac{1}{2})^2\bigr):1{,}35-\left(-\dfrac{3}{5}\right)^3 \cdot 10^3$$

$$-2 \dfrac{1}{2}=-\dfrac{5}{2}, \quad \left(-\dfrac{5}{2}\right)^2=\dfrac{25}{4}$$

$$7-\dfrac{25}{4}=\dfrac{28}{4}-\dfrac{25}{4}=\dfrac{3}{4}$$

$$1{,}35=\dfrac{135}{100}=\dfrac{27}{20}$$

$$\dfrac{3}{4}:\dfrac{27}{20}=\dfrac{3}{4}\cdot \dfrac{20}{27}=\dfrac{5}{9}$$

$$\left(-\dfrac{3}{5}\right)^3=-\dfrac{27}{125}$$

$$-\dfrac{27}{125}\cdot 10^3=-\dfrac{27}{125}\cdot 1000=-216$$

$$\dfrac{5}{9}-(-216)=216 \dfrac{5}{9}$$