Упр.15.6 ГДЗ Мордкович Семенов 8 класс (Алгебра)

а) (-(2x^3 y^4)/(5a^2 b))^3 · (-(25a^4 b^3)/(24x^8 y^13));
б) (-(10p^2 q^2)/(3a^3))^2 : (-(25p^3 q^3)/(27a^6));
в) ((-2a^8 b^3)/c^7 )^5 : (-(4a^10 b^4)/c^9)^4;
г) (x^2/(2a^3 ))^3 · (-(4a^4)/x^3)^2;
д) (-(9x^7 y^6)/a^12)^4 · (-a^8/(27x^5 y^4))^3;
е) (-(2a^2)/(9b^3))^6 : (-(4a^4)/(27b^5))^4.

1. а)

   $$\left(-\frac{2x^3y^4}{5a^2b}\right)^3\cdot\left(-\frac{25a^4b^3}{24x^8y^{13}}\right) =-\frac{8x^9y^{12}}{125a^6b^3}\cdot\left(-\frac{25a^4b^3}{24x^8y^{13}}\right)$$
   $$=\frac{8\cdot25\cdot x^9y^{12}a^4b^3}{125\cdot24\cdot a^6b^3x^8y^{13}} =\frac{x}{15a^2y}.$$

2. б)

   $$\left(-\frac{10p^2q^2}{3a^3}\right)^2:\left(-\frac{25p^3q^3}{27a^6}\right) =\frac{100p^4q^4}{9a^6}\cdot\left(-\frac{27a^6}{25p^3q^3}\right)$$
   $$=-\frac{100\cdot27}{9\cdot25}\,pq=-12pq.$$

3. в)

   $$\left(-\frac{2a^8b^3}{c^7}\right)^5:\left(-\frac{4a^{10}b^4}{c^9}\right)^4 =-\frac{32a^{40}b^{15}}{c^{35}}:\frac{256a^{40}b^{16}}{c^{36}}$$
   $$=-\frac{32a^{40}b^{15}}{c^{35}}\cdot\frac{c^{36}}{256a^{40}b^{16}} =-\frac{c}{8b}.$$

4. г)

   $$\left(\frac{x^2}{2a^3}\right)^3\cdot\left(-\frac{4a^4}{x^3}\right)^2 =\frac{x^6}{8a^9}\cdot\frac{16a^8}{x^6} =\frac{2}{a}.$$

5. д)

   $$\left(-\frac{9x^7y^6}{a^{12}}\right)^4\cdot\left(-\frac{a^8}{27x^5y^4}\right)^3 =\frac{3^8x^{28}y^{24}}{a^{48}}\cdot\left(-\frac{a^{24}}{3^9x^{15}y^{12}}\right)$$
   $$=-\frac{x^{13}y^{12}}{3a^{24}}.$$

6. е)

   $$\left(-\frac{2a^2}{9b^3}\right)^6:\left(-\frac{4a^4}{27b^5}\right)^4 =\frac{2^6a^{12}}{3^{12}b^{18}}:\frac{2^8a^{16}}{3^{12}b^{20}}$$
   $$=\frac{2^6a^{12}}{3^{12}b^{18}}\cdot\frac{3^{12}b^{20}}{2^8a^{16}} =\frac{b^2}{4a^4}.$$

Ответ

а) $$\frac{x}{15a^2y}$$; б) $$-12pq$$; в) $$-\frac{c}{8b}$$; г) $$\frac{2}{a}$$; д) $$-\frac{x^{13}y^{12}}{3a^{24}}$$; е) $$\frac{b^2}{4a^4}$$.