Упр.15.6 ГДЗ Мордкович Семенов 8 класс (Алгебра)
- Упростите выражение: а) $$\left(-\frac{2x^3y^4}{5a^2b}\right)^3\cdot\left(-\frac{25a^4b^3}{24x^8y^{13}}\right)$$; б) $$\left(-\frac{10p^2q^2}{3a^3}\right)^2:\left(-\frac{25p^3q^3}{27a^6}\right)$$; в) $$\left(\frac{-2a^8b^3}{c^7}\right)^5:\left(-\frac{4a^{10}b^4}{c^9}\right)^4$$; г) $$\left(\frac{x^2}{2a^3}\right)^3\cdot\left(-\frac{4a^4}{x^3}\right)^2$$; д) $$\left(-\frac{9x^7y^6}{a^{12}}\right)^4\cdot\left(-\frac{a^8}{27x^5y^4}\right)^3$$; е) $$\left(-\frac{2a^2}{9b^3}\right)^6:\left(-\frac{4a^4}{27b^5}\right)^4$$.
а)
$$\left(-\frac{2x^3y^4}{5a^2b}\right)^3\cdot\left(-\frac{25a^4b^3}{24x^8y^{13}}\right) =-\frac{8x^9y^{12}}{125a^6b^3}\cdot\left(-\frac{25a^4b^3}{24x^8y^{13}}\right)$$
$$=\frac{8\cdot25\cdot x^9y^{12}a^4b^3}{125\cdot24\cdot a^6b^3x^8y^{13}} =\frac{x}{15a^2y}.$$б)
$$\left(-\frac{10p^2q^2}{3a^3}\right)^2:\left(-\frac{25p^3q^3}{27a^6}\right) =\frac{100p^4q^4}{9a^6}\cdot\left(-\frac{27a^6}{25p^3q^3}\right)$$
$$=-\frac{100\cdot27}{9\cdot25}\,pq=-12pq.$$в)
$$\left(-\frac{2a^8b^3}{c^7}\right)^5:\left(-\frac{4a^{10}b^4}{c^9}\right)^4 =-\frac{32a^{40}b^{15}}{c^{35}}:\frac{256a^{40}b^{16}}{c^{36}}$$
$$=-\frac{32a^{40}b^{15}}{c^{35}}\cdot\frac{c^{36}}{256a^{40}b^{16}} =-\frac{c}{8b}.$$г)
$$\left(\frac{x^2}{2a^3}\right)^3\cdot\left(-\frac{4a^4}{x^3}\right)^2 =\frac{x^6}{8a^9}\cdot\frac{16a^8}{x^6} =\frac{2}{a}.$$
д)
$$\left(-\frac{9x^7y^6}{a^{12}}\right)^4\cdot\left(-\frac{a^8}{27x^5y^4}\right)^3 =\frac{3^8x^{28}y^{24}}{a^{48}}\cdot\left(-\frac{a^{24}}{3^9x^{15}y^{12}}\right)$$
$$=-\frac{x^{13}y^{12}}{3a^{24}}.$$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) $$\frac{b^2}{4a^4}$$.








