№ 21 ЗПС Алгебра = № 21 ЗПС Математика
Спростіть вираз:
1. $\frac{(\sqrt x+\sqrt y)^2-4y}{(x-y):\left(\frac{1}{\sqrt y}+\frac{3}{\sqrt x}\right)}:\frac{x+9y+6\sqrt{xy}}{\frac{1}{\sqrt x}+\frac{1}{\sqrt y}};$
2. $\frac{2\cdot\sqrt{1+\frac{1}{4}\left(\frac{1}{\sqrt a}-\sqrt a\right)^2}}{\sqrt{1+\frac{1}{4}\left(\frac{1}{\sqrt a}-\sqrt a\right)^2}-\frac{1}{2}\left(\frac{1}{\sqrt a}-\sqrt a\right)}.$
Розв'язок:
1. $\frac{(\sqrt x+\sqrt y)^2-4y}{(x-y):\left(\frac{1}{\sqrt y}+\frac{3}{\sqrt x}\right)}:$
$:\frac{x+9y+6\sqrt{xy}}{\frac{1}{\sqrt x}+\frac{1}{\sqrt y}}=\frac{1}{xy};$
а) $ \frac{(\sqrt x+\sqrt y)^2-4y}{(x-y):\left(\frac{1}{\sqrt y}+\frac{3}{\sqrt x}\right)}=$
$\frac{(\sqrt x+\sqrt y-2\sqrt y)(\sqrt x+\sqrt y+2\sqrt y)}{\frac{x-y}{1}\cdot\frac{\sqrt{xy}}{\sqrt x+3\sqrt y}}=$
$= \frac{(\sqrt x-\sqrt y)(\sqrt x+3\sqrt y)(\sqrt x+3\sqrt y)}{\sqrt{xy}\cdot(x-y)}=$
$= \frac{(\sqrt x-\sqrt y)\cdot(\sqrt x+3\sqrt y)^2}{\sqrt{xy}\cdot(\sqrt x-\sqrt y)(\sqrt x+\sqrt y)}=$
$= \frac{(\sqrt x+3\sqrt y)^2}{\sqrt{xy}\cdot(\sqrt x+\sqrt y)};$
б) $\frac{x+9y+6\sqrt{xy}}{\frac{1}{\sqrt x}+\frac{1}{\sqrt y}}= \frac{(\sqrt x+3\sqrt y)^2}{\frac{\sqrt x+\sqrt y}{\sqrt{xy}}}=$
$= \frac{\sqrt{xy}\cdot(\sqrt x+3\sqrt y)^2}{\sqrt x+\sqrt y};$
в) $\frac{(\sqrt x+3\sqrt y)^2}{\sqrt{xy}\cdot(\sqrt x+\sqrt y)}:\frac{\sqrt{xy}\cdot(\sqrt x+3\sqrt y)^2}{\sqrt x+\sqrt y}=$
$= \frac{(\sqrt x+3\sqrt y)^2\cdot(\sqrt x+\sqrt y)}{\sqrt{xy}\cdot(\sqrt x+\sqrt y)\cdot\sqrt{xy}\cdot(\sqrt x+3\sqrt y)^2}=$
$= \frac{1}{(\sqrt{xy})^2}=\frac{1}{xy}.$
2. $\frac{2\cdot\sqrt{1+\frac{1}{4}\left(\frac{1}{\sqrt a}-\sqrt a\right)^2}}{\sqrt{1+\frac{1}{4}\left(\frac{1}{\sqrt a}-\sqrt a\right)^2}-\frac{1}{2}\left(\frac{1}{\sqrt a}-\sqrt a\right)}=$
$= \frac{1+a}{a};$
a) $ 2\cdot\sqrt{1+\frac{1}{4}\left(\frac{1}{\sqrt a}-\sqrt a\right)^2}=$
$= 2\cdot\sqrt{1+\frac{1}{4}\left(\frac{1}{a}-2+a\right)}=$
$= 2\cdot\sqrt{1+\frac{1}{4a}-\frac{1}{2}+\frac{a}{4}}=$
$= 2\cdot\sqrt{\frac{1}{2}+\frac{1+a^2}{4a}}=$
$= 2\cdot\sqrt{\frac{2a+1+a^2}{4a}}=$
$= 2\cdot\sqrt{\frac{(a+1)^2}{4a}}=\frac{|a+1|}{\sqrt a}.$
Оскільки $a\geq0,$ то $\frac{|a+1|}{\sqrt a}=\frac{a+2}{\sqrt a};$
б) $\sqrt{1+\frac{1}{4}\left(\frac{1}{\sqrt a}-\sqrt a\right)^2}-$
$- \frac{1}{2}\left(\frac{1}{\sqrt a}-\sqrt a\right)=$
$= \frac{a+1}{2\sqrt a}-\frac{1}{2\sqrt a}+\frac{\sqrt a}{2}=$
$= \frac{a+1-1+a}{2\sqrt a}=\frac{2a}{2\sqrt a}=\frac{a}{\sqrt a};$
в) $\frac{a+1}{\sqrt a}:\frac{a}{\sqrt a}=$
$= \frac{a+1}{\sqrt a}\cdot\frac{\sqrt a}{a}=\frac{a+1}{a}.$