Giải câu 4 bài: Lũy thừa

Giải câu 4 bài: Lũy thừa.

a) $\frac{a^{\frac{4}{3}} (a^{- \frac{1}{3}} + a^{\frac{2}{3}})}{a^{\frac{1}{4}} (a^{\frac{3}{4}} + a^{- \frac{1}{4}})}$

= $\frac{a^{\frac{4}{3}} . a^{- \frac{1}{3}} + a^{\frac{4}{3}} . a^{\frac{2}{3}}}{a^{\frac{1}{4}} . a^{\frac{3}{4}} + a^{\frac{1}{4}} . a^{- \frac{1}{4}}}$

= $\frac{a^{\frac{4}{3} - \frac{1}{3}} + a^{\frac{4}{3} + \frac{2}{3}}}{a^{\frac{1}{4} + \frac{3}{4}} + a^{\frac{1}{4} - \frac{1}{4}}}$

= $\frac{a + a^{2}}{a + 1} = \frac{a (a + 1)}{a + 1} = a$

Vậy $\frac{a^{\frac{4}{3}} (a^{- \frac{1}{3}} + a^{\frac{2}{3}})}{a^{\frac{1}{4}} (a^{\frac{3}{4}} + a^{- \frac{1}{4}})} = a$

b) $\frac{b^{\frac{1}{5}} (\sqrt[5]{b^{4}} - \sqrt[5]{b^{- 1}})}{b^{\frac{2}{3}} (\sqrt[3]{b} - \sqrt[3]{b^{- 2}})}$

= $\frac{b^{\frac{1}{5}} . b^{\frac{1}{4}} - b^{\frac{1}{5}} . b^{- \frac{1}{5}}}{b^{\frac{2}{3}} . b^{\frac{1}{3}} - b^{\frac{2}{3}} . b^{- \frac{2}{3}}}$

= $\frac{b - 1}{b - 1} = 1$

Vậy $\frac{b^{\frac{1}{5}} (\sqrt[5]{b^{4}} - \sqrt[5]{b^{- 1}})}{b^{\frac{2}{3}} (\sqrt[3]{b} - \sqrt[3]{b^{- 2}})} = 1$

c) $\frac{a^{\frac{1}{3}} b^{- \frac{1}{3}} - a^{- \frac{1}{3}} b^{\frac{1}{3}}}{\sqrt[3]{a^{2}} - \sqrt[3]{b^{2}}}$

= $\frac{a^{\frac{2}{3} - \frac{1}{3}} b^{\frac{- 1}{3}} - a^{\frac{- 1}{2}} b^{\frac{2}{3} - \frac{1}{3}}}{a^{\frac{1}{6}} + a^{\frac{1}{6}}}$

= $(a b)^{\frac{- 1}{3}} = \frac{1}{\sqrt[3]{a b}}$

Vậy $\frac{a^{\frac{1}{3}} b^{- \frac{1}{3}} - a^{- \frac{1}{3}} b^{\frac{1}{3}}}{\sqrt[3]{a^{2}} - \sqrt[3]{b^{2}}} = \frac{1}{\sqrt[3]{a b}}$

d) $\frac{a^{\frac{1}{3}} \sqrt{b} + b^{\frac{1}{3}} \sqrt{a}}{\sqrt[6]{a} + \sqrt[6]{b}}$

= $\frac{a^{\frac{1}{3}} . b^{\frac{1}{2}} + b^{\frac{1}{3}} . a^{\frac{1}{2}}}{a^{\frac{1}{6}} + b^{\frac{1}{6}}}$

= $\frac{(a b)^{\frac{1}{3}} [b^{\frac{1}{6}} + a^{\frac{1}{6}}]}{a^{\frac{1}{6}} + b^{\frac{1}{6}}} = \sqrt[3]{a b}$

Vậy $\frac{a^{\frac{1}{3}} \sqrt{b} + b^{\frac{1}{3}} \sqrt{a}}{\sqrt[6]{a} + \sqrt[6]{b}} = \sqrt[3]{a b}$

Giải câu 4 bài: Lũy thừa

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