Question

16．化簡$\frac{y\sqrt{x}+x\sqrt{y}}{xy-{y}^{2}}$-$\frac{x+\sqrt{xy}+y}{x\sqrt{x}-y\sqrt{y}}$．

Answer 1

分析 根據(jù)平方差公式，立方差公式，有理數(shù)指數(shù)冪的運算性質(zhì)，可對已知的式子進行化簡．

解答 解：$\frac{y\sqrt{x}+x\sqrt{y}}{xy-{y}^{2}}$-$\frac{x+\sqrt{xy}+y}{x\sqrt{x}-y\sqrt{y}}$
=$\frac{\sqrt{xy}（\sqrt{y}+\sqrt{x}）}{y（x-y）}$-$\frac{x+\sqrt{xy}+y}{{\sqrt{x}}^{3}-{\sqrt{y}}^{3}}$
=$\frac{\sqrt{xy}（\sqrt{y}+\sqrt{x}）}{y（\sqrt{y}+\sqrt{x}）（-\sqrt{y}+\sqrt{x}）}$-$\frac{x+\sqrt{xy}+y}{{（\sqrt{x}}^{\;}-{\sqrt{y}}^{\;}）（x+\sqrt{xy}+y）}$
=$\frac{\sqrt{x}}{\sqrt{y}{（\sqrt{x}}^{\;}-{\sqrt{y}}^{\;}）}$-$\frac{1}{{（\sqrt{x}}^{\;}-{\sqrt{y}}^{\;}）}$
=$\frac{\sqrt{x}}{\sqrt{y}{（\sqrt{x}}^{\;}-{\sqrt{y}}^{\;}）}$-$\frac{\sqrt{y}}{\sqrt{y}{（\sqrt{x}}^{\;}-{\sqrt{y}}^{\;}）}$
=$\frac{\sqrt{x}-{\sqrt{y}}^{\;}}{\sqrt{y}{（\sqrt{x}}^{\;}-{\sqrt{y}}^{\;}）}$
=${\frac{1}{\sqrt{y}}}^{\;}$
=$\frac{\sqrt{y}}{y}$

點評 本題考查的知識點是有理數(shù)指數(shù)冪的運算性質(zhì)，熟練掌握相關(guān)公式及運算性質(zhì)是解答的關(guān)鍵．