Question

5．计算（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 根据平方差公式和除法法则进行化简即可解答本题．

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

点评 本题考查分数指数幂，解题的关键是巧妙的利用平方差公式化简．