Упр.878 ГДЗ Ткачёва 6 класс (Математика)
Найти значение выражения:
- 1) $$0{,}0075 \cdot 10^2 — 3{,}2 \cdot (0{,}08 — 0{,}2^2) + 40 : 10^3$$;
2) $$16{,}4 : 10^3 + 4{,}1 \cdot (3{,}25 — 2{,}12) — 0{,}86 \cdot 5^2$$;
3) $$\left(-1\frac{2}{7}\right)^2 \cdot \frac{7}{27} — 0{,}6 : 1\frac{1}{2} — \left(-1\frac{1}{3}\right)^3$$;
4) $$\left(7 — \left(-2\frac{1}{2}\right)^2\right) : 1{,}35 — \left(-\frac{3}{5}\right)^3 \cdot 10^3$$.
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) $$\bigl(7-(-2 \dfrac{1}{2})^2\bigr):1{,}35-\left(-\dfrac{3}{5}\right)^3 \cdot 10^3$$
$$-2 \dfrac{1}{2}=-\dfrac{5}{2}, \quad \left(-\dfrac{5}{2}\right)^2=\dfrac{25}{4}$$
$$7-\dfrac{25}{4}=\dfrac{28}{4}-\dfrac{25}{4}=\dfrac{3}{4}$$
$$1{,}35=\dfrac{135}{100}=\dfrac{27}{20}$$
$$\dfrac{3}{4}:\dfrac{27}{20}=\dfrac{3}{4}\cdot \dfrac{20}{27}=\dfrac{5}{9}$$
$$\left(-\dfrac{3}{5}\right)^3=-\dfrac{27}{125}$$
$$-\dfrac{27}{125}\cdot 10^3=-\dfrac{27}{125}\cdot 1000=-216$$
$$\dfrac{5}{9}-(-216)=216 \dfrac{5}{9}$$
Ответ
1) $$0{,}662$$; 2) $$-26{,}2396$$; 3) $$2 \dfrac{377}{945}$$; 4) $$216 \dfrac{5}{9}$$.
