Question

13．化簡：$\frac{m-{m}^{-1}}{{m}^{\frac{2}{3}}-{m}^{-\frac{2}{3}}}$-$\frac{m+{m}^{-1}}{{m}^{\frac{2}{3}}+2+{m}^{-\frac{2}{3}}}$+$\frac{2m}{{m}^{\frac{2}{3}}+1}$（m＞0）

Answer 1

分析 根據(jù)完全平方公式，平方差公式，立方和公式，立方差公式及對數(shù)的運(yùn)算性質(zhì)，可化簡原式．

解答 解：：$\frac{m-{m}^{-1}}{{m}^{\frac{2}{3}}-{m}^{-\frac{2}{3}}}$-$\frac{m+{m}^{-1}}{{m}^{\frac{2}{3}}+2+{m}^{-\frac{2}{3}}}$+$\frac{2m}{{m}^{\frac{2}{3}}+1}$
=$\frac{{（m}^{\frac{1}{3}}-{m}^{-\frac{1}{3}}）（{m}^{\frac{2}{3}}+1+{m}^{-\frac{2}{3}}）}{{（m}^{\frac{1}{3}}-{m}^{-\frac{1}{3}}）{（m}^{\frac{1}{3}}+{m}^{-\frac{1}{3}}）}$-$\frac{{（m}^{\frac{1}{3}}+{m}^{-\frac{1}{3}}）（{m}^{\frac{2}{3}}-1+{m}^{-\frac{2}{3}}）}{{（m}^{\frac{1}{3}}+{m}^{-\frac{1}{3}}）^{2}}$+$\frac{2m}{{m}^{\frac{2}{3}}+1}$
=$\frac{（{m}^{\frac{2}{3}}+1+{m}^{-\frac{2}{3}}）}{{（m}^{\frac{1}{3}}+{m}^{-\frac{1}{3}}）}$-$\frac{（{m}^{\frac{2}{3}}-1+{m}^{-\frac{2}{3}}）}{{（m}^{\frac{1}{3}}+{m}^{-\frac{1}{3}}）}$+$\frac{2m}{{m}^{\frac{2}{3}}+1}$
=$\frac{2}{{（m}^{\frac{1}{3}}+{m}^{-\frac{1}{3}}）}$+$\frac{2m}{{m}^{\frac{2}{3}}+1}$
=$\frac{2{m}^{\frac{1}{3}}}{{m}^{\frac{2}{3}}+1}$+$\frac{2m}{{m}^{\frac{2}{3}}+1}$
=$\frac{2{m}^{\frac{1}{3}}（{m}^{\frac{2}{3}}+1）}{{m}^{\frac{2}{3}}+1}$
=$2{m}^{\frac{1}{3}}$

點(diǎn)評 本題考查的知識點(diǎn)是有理數(shù)指數(shù)冪的化簡求值，熟練掌握乘法公式，是解答的關(guān)鍵．