Упр.8.20 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1) 2a^3•(-5a^4 b^5 )^2;
2) (-x^6 y)^3•11x^4 y^5;
3) (-0,6a^3 b^5 c^6 )^2•3a^2 c^8;
4)-1 3/11 m^4 n^9•(-1/7 mn^3 )^2;
5) 1 7/9 x^7 y^2•(3/4 x^2 y^9 )^4;
6)-(-2c^2 d^5 )^7•(-1/2 c^4 d^5 )^4.

1. $$2a^3\cdot(-5a^4b^5)^2=2a^3\cdot25a^8b^{10}=50a^{11}b^{10}.$$
2. $$(-x^6y)^3\cdot11x^4y^5=-x^{18}y^3\cdot11x^4y^5=-11x^{22}y^8.$$
3. $$(-0{,}6a^3b^5c^6)^2\cdot3a^2c^8=0{,}36a^6b^{10}c^{12}\cdot3a^2c^8=1{,}08a^8b^{10}c^{20}.$$
4. $$-1\frac{3}{11}m^4n^9\cdot\left(-\frac{1}{7}mn^3\right)^2=-\frac{14}{11}m^4n^9\cdot\frac{1}{49}m^2n^6$$
   $$=-\frac{14}{11\cdot49}m^6n^{15}=-\frac{2}{77}m^6n^{15}.$$
5. $$1\frac{7}{9}x^7y^2\cdot\left(\frac{3}{4}x^2y^9\right)^4=\frac{16}{9}x^7y^2\cdot\frac{81}{256}x^8y^{36}$$
   $$=\frac{16\cdot81}{9\cdot256}x^{15}y^{38}=\frac{9}{16}x^{15}y^{38}.$$
6. $$-(-2c^2d^5)^7\cdot\left(-\frac{1}{2}c^4d^5\right)^4=-(-128c^{14}d^{35})\cdot\frac{1}{16}c^{16}d^{20}$$
   $$=128c^{14}d^{35}\cdot\frac{1}{16}c^{16}d^{20}=8c^{30}d^{55}.$$

Ответ

1) $$50a^{11}b^{10}$$; 2) $$-11x^{22}y^8$$; 3) $$1{,}08a^8b^{10}c^{20}$$; 4) $$-\frac{2}{77}m^6n^{15}$$; 5) $$\frac{9}{16}x^{15}y^{38}$$; 6) $$8c^{30}d^{55}$$.