Question

5．計(jì)算（a${\;}^{\frac{4}{5}}$-b${\;}^{\frac{4}{5}}$）÷（a${\;}^{\frac{2}{5}}$+b${\;}^{\frac{2}{5}}$）÷（a${\;}^{\frac{1}{5}}$+b${\;}^{\frac{1}{5}}$）

Answer 1

分析 根據(jù)平方差公式和除法法則進(jìn)行化簡(jiǎn)即可解答本題．

解答 解：（a${\;}^{\frac{4}{5}}$-b${\;}^{\frac{4}{5}}$）÷（a${\;}^{\frac{2}{5}}$+b${\;}^{\frac{2}{5}}$）÷（a${\;}^{\frac{1}{5}}$+b${\;}^{\frac{1}{5}}$）
=$（{a}^{\frac{2}{5}}+^{\frac{2}{5}}）（{a}^{\frac{2}{5}}-^{\frac{2}{5}}）$×$\frac{1}{{a}^{\frac{2}{5}}+^{-\frac{2}{5}}}×\frac{1}{{a}^{\frac{1}{5}}+^{\frac{1}{5}}}$
=$（{a}^{\frac{2}{5}}-^{\frac{2}{5}}）×\frac{1}{{a}^{\frac{1}{5}}+^{\frac{1}{5}}}$
=$（{a}^{\frac{1}{5}}+^{\frac{1}{5}}）（{a}^{\frac{1}{5}}-^{\frac{1}{5}}）×\frac{1}{{a}^{\frac{1}{5}}+^{\frac{1}{5}}}$
=${a}^{\frac{1}{5}}-^{\frac{1}{5}}$．

點(diǎn)評(píng) 本題考查分?jǐn)?shù)指數(shù)冪，解題的關(guān)鍵是巧妙的利用平方差公式化簡(jiǎn)．