Question

giúp em với ạ

Answer 1

ĐKXĐ: x>0; y>0; x<>y

TA có: \(\frac{\sqrt{x}-\sqrt{y}}{x\sqrt{y}+y\sqrt{x}}+\frac{\sqrt{x}+\sqrt{y}}{x\sqrt{y}-y\cdot\sqrt{x}}\)

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

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

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

\(P=\left(\frac{\sqrt{x}-\sqrt{y}}{x\sqrt{y}+y\sqrt{x}}+\frac{\sqrt{x}+\sqrt{y}}{x\sqrt{y}-y\cdot\sqrt{x}}\right)\cdot\frac{\sqrt{x^3y}}{x+y}-\frac{2y}{x-y}\)

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

=>P là số nguyên với mọi x,y thỏa mãn ĐKXĐ