Question

допоможіть спростити вираз: ​

Answer 1

$\displaystyle (\frac{x^{0,5}+3y^{0,5}}{x-2x^{0,5}y^{0,5}+y}+\frac{x^{0,5}-3y^{0,5}}{x-y})*\frac{x^{0,5}-y^{0,5}}{2} = \\ =(\frac{\sqrt{x} +3\sqrt{y} }{x-2\sqrt{xy}+y}+\frac{\sqrt{x} -3\sqrt{y} }{x-y})*\frac{\sqrt{x} -\sqrt{y} }{2} = \\ = (\frac{\sqrt{x} +3\sqrt{y} }{(\sqrt{x} -\sqrt{y} )^{2}}+\frac{\sqrt{x} -3\sqrt{y} }{(\sqrt{x})^{2}-(\sqrt{y} )^{2}})*\frac{\sqrt{x} -\sqrt{y} }{2} =$
$\displaystyle = \frac{\sqrt{x} +3\sqrt{y} }{(\sqrt{x} -\sqrt{y} )^{2}} *\frac{\sqrt{x} -\sqrt{y} }{2} +\frac{\sqrt{x} -3\sqrt{y} }{(\sqrt{x} -\sqrt{y} )(\sqrt{x} +\sqrt{y} )} *\frac{\sqrt{x} -\sqrt{y} }{2} = \\ = \frac{\sqrt{x} +3\sqrt{y} }{2(\sqrt{x} -\sqrt{y} )}+\frac{\sqrt{x} -3\sqrt{y} }{2(\sqrt{x} +\sqrt{y} )} = \frac{(\sqrt{x} +3\sqrt{y} )(\sqrt{x} +\sqrt{y} )+(\sqrt{x} -3\sqrt{y} )(\sqrt{x} -\sqrt{y} )}{2(x+y)} = \\ = \frac{x+\sqrt{xy}+3\sqrt{xy}+3y+x-\sqrt{xy}-3\sqrt{xy}+3y}{2(x+y)}=$
$\displaystyle =\frac{2x+6y}{2(x+y)} =\frac{x+3y}{x+y}$