Question

rút gọn A

A=\(\left(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right)\)*\(\frac{1}{x-y}+\frac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

Answer 1

ĐKXĐ: \(x\ge0;y\ge0;x\ne y\)

A = \(\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right).\frac{1}{x-y}\)+\(\frac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

= \(\left(x-2\sqrt{xy}+y\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)+\(\frac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

=\(\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}+\frac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}+\frac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

=\(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}=1\)

Vậy A = 1