Упр.8.21 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1) 20a^8•(9a)^2;
2) (-b^5 )^4•12b^6;
3) (3m^6 n^3 )^4•(-1/81 m^9 n);
4) (0,2x^7 y^8 )^3•6x^2 y^2;
5) (-1/2 ab^4 )^3•(4a^6 )^2;
6) (-2/3 x^2 y)^5•(-3/4 xy^2 )^2.

1. $$20a^8 \cdot (9a)^2 = 20a^8 \cdot 81a^2 = 1620a^{10}.$$
2. $$(-b^5)^4 \cdot 12b^6 = b^{20} \cdot 12b^6 = 12b^{26}.$$
3. $$ (3m^6n^3)^4 \cdot \left(-\frac{1}{81}m^9n\right) = 81m^{24}n^{12} \cdot \left(-\frac{1}{81}m^9n\right) = -m^{33}n^{13}. $$
4. $$ (0{,}2x^7y^8)^3 \cdot 6x^2y^2 = 0{,}008x^{21}y^{24} \cdot 6x^2y^2 = 0{,}048x^{23}y^{26}. $$
5. $$ \left(-\frac{1}{2}ab^4\right)^3 \cdot (4a^6)^2 = -\frac{1}{8}a^3b^{12} \cdot 16a^{12} = -2a^{15}b^{12}. $$
6. $$ \left(-\frac{2}{3}x^2y\right)^5 \cdot \left(-\frac{3}{4}xy^2\right)^2 = -\frac{32}{243}x^{10}y^5 \cdot \frac{9}{16}x^2y^4 = -\frac{2}{27}x^{12}y^9. $$

Ответ

1) $$1620a^{10}$$; 2) $$12b^{26}$$; 3) $$-m^{33}n^{13}$$; 4) $$0{,}048x^{23}y^{26}$$; 5) $$-2a^{15}b^{12}$$; 6) $$-\frac{2}{27}x^{12}y^9$$.