Kết quả rút gọn biểu thức \(A=\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\) là
\(A=1\)\(A=\sqrt{x}+\sqrt{y}\)\(A=\sqrt{x}-\sqrt{y}\)\(A=2\sqrt{y}\)Hướng dẫn giải:Điều kiện: \(x\ge0;y\ge0;x+y>0\)
\(A=\left(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x^2}-\sqrt{xy}+\sqrt{y^2}\right)}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\\ =\left(\sqrt{x^2}-2\sqrt{xy}+\sqrt{y^2}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\\ =\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}=1\)
