Упр.68 ГДЗ Колягин Ткачёва 9 класс (Алгебра)
Упростить выражение:
- 1) $$\frac{a^{\frac{4}{3}}\left(a^{-\frac{1}{3}}+a^{\frac{2}{3}}\right)}{a^{\frac{1}{4}}\left(a^{\frac{3}{4}}+a^{-\frac{1}{4}}\right)}$$;
2) $$\frac{b^{\frac{1}{5}}\left((b^4)^{\frac{1}{5}}-(b^{-1})^{\frac{1}{5}}\right)}{b^{\frac{2}{3}}\left(b^{\frac{1}{3}}-(b^{-2})^{\frac{1}{3}}\right)}$$;
3) $$\frac{a^{\frac{5}{3}}b^{-1}-ab^{-\frac{1}{3}}}{(a^2)^{\frac{1}{3}}-(b^2)^{\frac{1}{3}}}$$;
4) $$\frac{a^{\frac{1}{3}}\sqrt{b}+b^{\frac{1}{3}}\sqrt{a}}{a^{\frac{1}{6}}+b^{\frac{1}{6}}}$$.
$$\frac{a^{4/3}\left(a^{-1/3}+a^{2/3}\right)}{a^{1/4}\left(a^{3/4}+a^{-1/4}\right)}= \frac{a^{4/3}\cdot a^{-1/3}+a^{4/3}\cdot a^{2/3}}{a^{1/4}\cdot a^{3/4}+a^{1/4}\cdot a^{-1/4}}= \frac{a+a^2}{a+1}$$
$$\frac{a+a^2}{a+1}=\frac{a(1+a)}{a+1}=a$$
$$\frac{b^{1/5}\left((b^4)^{1/5}-(b^{-1})^{1/5}\right)}{b^{2/3}\left(b^{1/3}-(b^{-2})^{1/3}\right)}= \frac{b^{1/5}\left(b^{4/5}-b^{-1/5}\right)}{b^{2/3}\left(b^{1/3}-b^{-2/3}\right)}$$
$$=\frac{b^{1}-b^{0}}{b^{1}-b^{0}}=1$$
$$\frac{a^{5/3}b^{-1}-ab^{-1/3}}{(a^2)^{1/3}-(b^2)^{1/3}}= \frac{ab^{-1}\left(a^{2/3}-b^{2/3}\right)}{a^{2/3}-b^{2/3}}=ab^{-1}=\frac{a}{b}$$
$$\frac{a^{1/3}\sqrt{b}+b^{1/3}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}= \frac{a^{1/3}b^{1/2}+b^{1/3}a^{1/2}}{a^{1/6}+b^{1/6}}$$
$$=\frac{a^{1/3}b^{1/3}\left(b^{1/6}+a^{1/6}\right)}{a^{1/6}+b^{1/6}}=a^{1/3}b^{1/3}=\sqrt[3]{ab}$$
Ответ
1) $$a$$; 2) $$1$$; 3) $$\frac{a}{b}$$; 4) $$\sqrt[3]{ab}$$.
