Упр.8.10 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1) 12a^2•5a^3 b^7;
2)-4m^3•0,25m^6;
3) 3ab•(-17a^2 b);
4) 56x^5 y^14•2/7 x^2 y;
5)-1/3 p^2•(-27k)•5pk;
6) 2 1/4 b^2 c^5 d^3•(-3 1/3 b^3 c^4 d^7 ).

1. $$12a^2 \cdot 5a^3b^7 = 60a^{2+3}b^7 = 60a^5b^7.$$
2. $$-4m^3 \cdot 0{,}25m^6 = -1 \cdot m^{3+6} = -m^9.$$
3. $$3ab \cdot (-17a^2b) = -51a^{1+2}b^{1+1} = -51a^3b^2.$$
4. $$56x^5y^{14} \cdot \frac{2}{7}x^2y = \frac{56 \cdot 2}{7}x^{5+2}y^{14+1} = 16x^7y^{15}.$$
5. $$-\frac{1}{3}p^2 \cdot (-27k) \cdot 5pk = \frac{1}{3}\cdot 27 \cdot 5 \, p^{2+1}k^{1+1} = 45p^3k^2.$$
6. $$2\frac{1}{4}b^2c^5d^3 \cdot \left(-3\frac{1}{3}b^3c^4d^7\right) = \frac{9}{4}b^2c^5d^3 \cdot \left(-\frac{10}{3}b^3c^4d^7\right)$$
   $$= -\frac{9}{4}\cdot \frac{10}{3} \, b^{2+3}c^{5+4}d^{3+7} = -\frac{15}{2}b^5c^9d^{10} = -7{,}5b^5c^9d^{10}.$$

Ответ

1) $$60a^5b^7$$; 2) $$-m^9$$; 3) $$-51a^3b^2$$; 4) $$16x^7y^{15}$$; 5) $$45p^3k^2$$; 6) $$-7{,}5b^5c^9d^{10}$$.