Question

Rút gọn biểu thức:

A = \(\left[\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right].\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{x}+\dfrac{1}{y}:\dfrac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+y^3}{\sqrt{x^3y}+\sqrt{xy^3}}\)

Answer 1

\(ĐKXĐ:xy>0\)

\(A=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\cdot\dfrac{2}{\sqrt{x}+\sqrt{y}}+\dfrac{x+y}{xy}:\dfrac{\sqrt{x}\left(x+y\right)+\sqrt{y}\left(x+y\right)}{\sqrt{xy}\left(x+y\right)}\\ A=\dfrac{2}{\sqrt{xy}}+\dfrac{x+y}{xy}\cdot\dfrac{\sqrt{xy}\left(x+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(x+y\right)}\\ A=\dfrac{2}{\sqrt{xy}}+\dfrac{x+y}{xy}\cdot\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\\ A=\dfrac{2}{\sqrt{xy}}+\dfrac{x+y}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\\ A=\dfrac{2\sqrt{x}+2\sqrt{y}+x+y}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\)