Упр.42.29 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)
- Упростите выражение:
1) $$\left(\frac{81a^{-\frac{2}{3}}b^{-\frac{1}{3}}}{16c^{-\frac{4}{3}}}\right)^{-\frac{3}{4}}$$;
2) $$\left(\frac{125a^{-\frac{3}{2}}b^{-\frac{3}{2}}}{64c^{-\frac{1}{2}}}\right)^{-\frac{4}{3}}$$;
3) $$\left(\frac{a^{\frac{1}{2}}b^{\frac{1}{3}}+a^{\frac{1}{3}}b^{\frac{1}{2}}}{a^{\frac{1}{6}}+b^{\frac{1}{6}}}\right)^{-6}$$;
4) $$\left(\frac{a^{\frac{1}{4}}b^{\frac{1}{4}}-b^{\frac{1}{2}}}{a^{\frac{1}{2}}-a^{\frac{1}{4}}b^{\frac{1}{4}}}\right)^{-4}$$.
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) $$\left(\frac{a^{1/4}b^{1/4}-b^{1/2}}{a^{1/2}-a^{1/4}b^{1/4}}\right)^{-4} =\left(\frac{b^{1/4}\left(a^{1/4}-b^{1/4}\right)}{a^{1/4}\left(a^{1/4}-b^{1/4}\right)}\right)^{-4}$$
$$=\left(\frac{b^{1/4}}{a^{1/4}}\right)^{-4} =\left(\frac{b}{a}\right)^{-1} =\frac{a}{b}.$$
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) $$\frac{a}{b}$$.
