Упр.15.5 ГДЗ Мордкович Семенов 8 класс (Алгебра)

а) a^2/x · (x^2/a^3)^2;
б) (p/x^3)^3 : (p^2/x^3)^2;
в) ((x^6 y^8)/z^5 )^5 : (x^10 y^13)/z^8;
г) ((a^3 b)/c^4)^5 · (c^7/(a^5 b^2))^3;
д) -(50a^4 b^5)/(63m^9 n^8) : ((5a^2 b^3)/(3m^2 n^5))^3;
е) (-(2pq^5)/(3ma^2))^2 · (9m^2 a^2)/(4p^3 q^7).

1. а)

   $$\frac{a^2}{x}\cdot\left(\frac{x^2}{a^3}\right)^2 =\frac{a^2}{x}\cdot\frac{x^4}{a^6} =\frac{a^2x^4}{xa^6} =\frac{x^3}{a^4}.$$

2. б)

   $$\left(\frac{p}{x^3}\right)^3:\left(\frac{p^2}{x^3}\right)^2 =\frac{p^3}{x^9}:\frac{p^4}{x^6} =\frac{p^3}{x^9}\cdot\frac{x^6}{p^4} =\frac{1}{x^3p}.$$

3. в)

   $$\left(\frac{x^6y^8}{z^5}\right)^5:\frac{x^{10}y^{13}}{z^8} =\frac{x^{30}y^{40}}{z^{25}}:\frac{x^{10}y^{13}}{z^8} =\frac{x^{30}y^{40}}{z^{25}}\cdot\frac{z^8}{x^{10}y^{13}} =\frac{x^{20}y^{27}}{z^{17}}.$$

4. г)

   $$\left(\frac{a^3b}{c^4}\right)^5\cdot\left(\frac{c^7}{a^5b^2}\right)^3 =\frac{a^{15}b^5}{c^{20}}\cdot\frac{c^{21}}{a^{15}b^6} =\frac{c}{b}.$$

5. д)

   $$-\frac{50a^4b^5}{63m^9n^8}:\left(\frac{5a^2b^3}{3m^2n^5}\right)^3 =-\frac{50a^4b^5}{63m^9n^8}:\frac{125a^6b^9}{27m^6n^{15}}$$
   $$=-\frac{50a^4b^5}{63m^9n^8}\cdot\frac{27m^6n^{15}}{125a^6b^9} =-\frac{2\cdot3^7}{7\cdot5^2}\cdot\frac{n^7}{a^2b^4m^3} =-\frac{6n^7}{35a^2b^4m^3}.$$

6. е)

   $$\left(-\frac{2pq^5}{3ma^2}\right)^2\cdot\frac{9m^2a^2}{4p^3q^7} =\frac{4p^2q^{10}}{9m^2a^4}\cdot\frac{9m^2a^2}{4p^3q^7} =\frac{q^3}{a^2p}.$$

Ответ

а) $$\frac{x^3}{a^4}$$; б) $$\frac{1}{x^3p}$$; в) $$\frac{x^{20}y^{27}}{z^{17}}$$; г) $$\frac{c}{b}$$; д) $$-\frac{6n^7}{35a^2b^4m^3}$$; е) $$\frac{q^3}{a^2p}$$.