Question

15．化簡：a${\;}^{\frac{1}{3}}$+（a${\;}^{\frac{1}{3}}$-2b${\;}^{\frac{1}{2}}$）÷（a${\;}^{-\frac{2}{3}}$-$\frac{2\root{3}}{a}$）×$\frac{\sqrt{a•\root{3}{{a}^{2}}}}{\root{5}{\sqrt{a}•\root{3}{a}}}$．

Answer 1

分析 利用分?jǐn)?shù)指數(shù)冪的性質(zhì)和運(yùn)算法則直接求解即可．

解答 解：a${\;}^{\frac{1}{3}}$×（a${\;}^{\frac{1}{3}}$-2b${\;}^{\frac{1}{3}}$）÷（a${\;}^{-\frac{2}{3}}$-$\frac{2\root{3}}{a}$）×$\frac{\sqrt{a•\root{3}{{a}^{2}}}}{\root{5}{\sqrt{a}•\root{3}{a}}}$
=${a}^{\frac{1}{3}}$×$\frac{{a}^{\frac{1}{3}}-2^{\frac{1}{3}}}{{a}^{-\frac{2}{3}}-2{a}^{-1}^{\frac{1}{3}}}$×$\frac{{a}^{\frac{1}{2}}•{a}^{\frac{1}{3}}}{{a}^{\frac{1}{10}}•{a}^{\frac{1}{15}}}$
=${a}^{\frac{1}{3}}$×$\frac{{a}^{\frac{4}{3}}-2a^{\frac{1}{3}}}{{a}^{\frac{1}{3}}-2^{\frac{1}{3}}}$×a${\;}^{\frac{2}{3}}$
=a²．

點評 本題考查有理數(shù)指數(shù)冪的化簡求值，是基礎(chǔ)題，解題時要認(rèn)真審題，注意分?jǐn)?shù)指數(shù)冪的性質(zhì)和運(yùn)算法則的合理運(yùn)用．