Упр.38.13 ГДЗ Мерзляк Поляков 7 класс (Алгебра)

1) (33m^8)/(34n^8 ) : (88m^4)/(51n^4 ) : (21m^6)/(16n^2 );
2) ((2a^5)/y^6 )^4 : ((4a^6)/y^8 )^3;
3) (-(27x^3)/(16y^5 ))^2·((8y^3)/(9x^2 ))^3.

1. $$\frac{33m^8}{34n^8}:\frac{88m^4}{51n^4}:\frac{21m^6}{16n^2}= \frac{33m^8}{34n^8}\cdot\frac{51n^4}{88m^4}\cdot\frac{16n^2}{21m^6}$$

   $$=\frac{33\cdot 51\cdot 16\cdot m^8n^6}{34\cdot 88\cdot 21\cdot m^{10}n^8} =\frac{33\cdot 51\cdot 16}{34\cdot 88\cdot 21}\cdot\frac{1}{m^2n^2}$$

   $$\frac{33\cdot 51\cdot 16}{34\cdot 88\cdot 21}=\frac{3}{7}$$

   $$\frac{3}{7m^2n^2}$$

2. $$\left(\frac{2a^5}{y^6}\right)^4:\left(\frac{4a^6}{y^8}\right)^3= \frac{16a^{20}}{y^{24}}:\frac{64a^{18}}{y^{24}}$$

   $$=\frac{16a^{20}}{y^{24}}\cdot\frac{y^{24}}{64a^{18}}=\frac{a^2}{4}$$

3. $$\left(-\frac{27x^3}{16y^5}\right)^2\cdot\left(\frac{8y^3}{9x^2}\right)^3= \frac{(27x^3)^2}{(16y^5)^2}\cdot\frac{(8y^3)^3}{(9x^2)^3}$$

   $$=\frac{3^6x^6}{2^8y^{10}}\cdot\frac{2^9y^9}{3^6x^6} =\frac{2}{y}$$

Ответ

1) $$\frac{3}{7m^2n^2}$$; 2) $$\frac{a^2}{4}$$; 3) $$\frac{2}{y}$$.