Упр.42.29 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) ((81a^(-2/3)b^(-1/3))/(16c^(-4/3)))^(-3/4);
2) ((125a^(-3/2)b^(-3/2))/(64c^(-1/2)))^(-4/3);
3) ((a^(1/2)b^(1/3)+a^(1/3)b^(1/2))/(a^(1/6)+b^(1/6)))^(-6);
4) ((a^(1/4)b^(1/4)-b^(1/2))/(a^(1/2)-a^(1/4)b^(1/4)))^(-4).

1. $$\left(\frac{81a^{-2/3}b^{-1/3}}{16c^{-4/3}}\right)^{-3/4} =\left(\frac{3^4}{2^4}\cdot a^{-2/3}b^{-1/3}c^{4/3}\right)^{-3/4}$$
   $$=\left(\frac{3}{2}\right)^{-3}\cdot a^{1/2}b^{1/4}c^{-1} =\frac{8a^{1/2}b^{1/4}}{27c}.$$

2. $$\left(\frac{125a^{-3/2}b^{-3/2}}{64c^{-1/2}}\right)^{-4/3} =\left(\frac{5^3}{4^3}\cdot a^{-3/2}b^{-3/2}c^{1/2}\right)^{-4/3}$$
   $$=\left(\frac{5}{4}\right)^{-4}\cdot a^2b^2c^{-2/3} =\frac{256a^2b^2}{625c^{2/3}}.$$

3. $$\left(\frac{a^{1/2}b^{1/3}+a^{1/3}b^{1/2}}{a^{1/6}+b^{1/6}}\right)^{-6} =\left(\frac{a^{1/3}b^{1/3}\left(a^{1/6}+b^{1/6}\right)}{a^{1/6}+b^{1/6}}\right)^{-6}$$
   $$=\left(a^{1/3}b^{1/3}\right)^{-6} =(a^2b^2)^{-1} =\frac{1}{a^2b^2}.$$

4. $$\left(\frac{a^{1/4}b^{1/4}-b^{1/2}}{a^{1/2}-a^{1/4}b^{1/4}}\right)^{-4} =\left(\frac{b^{1/4}\left(a^{1/4}-b^{1/4}\right)}{a^{1/4}\left(a^{1/4}-b^{1/4}\right)}\right)^{-4}$$
   $$=\left(\frac{b}{a}\right)^{-1} =\frac{a}{b}.$$

Ответ

1) $$\frac{8a^{1/2}b^{1/4}}{27c}$$; 2) $$\frac{256a^2b^2}{625c^{2/3}}$$; 3) $$\frac{1}{a^2b^2}$$; 4) $$\frac{a}{b}$$.