№ 22.25 Алгебра = № 42.25 Математика
Спростіть вираз:
$\left(\frac{\sqrt x+\sqrt y}{\sqrt x}-\frac{\sqrt x}{\sqrt x-\sqrt y}\right)\cdot\frac{\sqrt x-\sqrt y}{\sqrt y}.$
Розв'язок:
$\left(\frac{\sqrt x+\sqrt y}{\sqrt x}-\frac{\sqrt x}{\sqrt x-\sqrt y}\right)\cdot\frac{\sqrt x-\sqrt y}{\sqrt y}=$
$= -\sqrt{\frac{y}{x}};$
a) $\frac{\sqrt x+\sqrt y}{\sqrt x}-\frac{\sqrt x}{\sqrt x-\sqrt y}=$
$= \frac{(\sqrt x-\sqrt y)(\sqrt x+\sqrt y)-(\sqrt x)^2}{\sqrt x\cdot\left(\sqrt x-\sqrt y\right)}=$
$= \frac{(\sqrt x)^2-(\sqrt y)^2-(\sqrt x)^2}{\sqrt x\left(\sqrt x-\sqrt y\right)}=$
$= -\frac{y}{\sqrt x\left(\sqrt x-\sqrt y\right)};$
б) $-\frac{y}{\sqrt x\left(\sqrt x-\sqrt y\right)}\cdot\frac{\sqrt x-\sqrt y}{\sqrt y}=$
$= -\frac{y\cdot\left(\sqrt x-\sqrt y\right)}{\sqrt x\left(\sqrt x-\sqrt y\right)\cdot\sqrt y}=$
$= -\frac{y}{\sqrt x\cdot\sqrt y}=-\frac{y\sqrt y}{\sqrt x(\sqrt y)^2}=$
$= -\frac{y\sqrt y}{y\sqrt x}=-\sqrt{\frac{y}{x}}.$