№ 19 ЗПС Алгебра = № 19 ЗПС Математика
Звільніться від ірраціональності в знаменнику дробу:
1. $\frac{\sqrt{3\sqrt2-2\sqrt3}}{\sqrt{3\sqrt2+2\sqrt3}};$
2. $\frac{(1+\sqrt3)^2-7}{\sqrt7+\sqrt3+1};$
3. $\frac{2}{\sqrt2+\sqrt{6+4\sqrt2}};$
4. $\frac{2+\sqrt3}{\sqrt6-\sqrt3+\sqrt2-1}.$
Розв'язок:
1. $\frac{\sqrt{3\sqrt2-2\sqrt3}}{\sqrt{3\sqrt2+2\sqrt3}}=$
$= \frac{\sqrt{(3\sqrt2-2\sqrt3)^2}}{\sqrt{(3\sqrt2+2\sqrt3)(3\sqrt2-2\sqrt3)}}=$
$= \frac{|3\sqrt2-2\sqrt3|}{\sqrt{(3\sqrt2)^2-(2\sqrt3)^2}}=$
$= \frac{3\sqrt2-2\sqrt3}{\sqrt{18-12}}= \frac{3\sqrt2-2\sqrt3}{\sqrt6}=$
$= \frac{\sqrt6(3\sqrt2-2\sqrt3)}{6}= \frac{3\sqrt{12}-2\sqrt{18}}{6}=$
$= \frac{3\sqrt{3\cdot4}-2\sqrt{2\cdot9}}{6}= \frac{6\sqrt3-6\sqrt2}{6}=$
$= \frac{6(\sqrt3-\sqrt2)}{6}=\sqrt3-\sqrt2;$
2. $\frac{(1+\sqrt3)^2-7}{\sqrt7+\sqrt3+1}=$
$= \frac{\left(1+\sqrt3-\sqrt7\right)\left(1+\sqrt3+\sqrt7\right)}{\sqrt7+\sqrt3+1}=$
$= 1+\sqrt3-\sqrt7;$
3. $\frac{2}{\sqrt2+\sqrt{6+4\sqrt2}}=$
$= \frac{2\left(\sqrt2-\sqrt{6+4\sqrt2}\right)}{\left(\sqrt2+\sqrt{6+4\sqrt2}\right)\left(\sqrt2-\sqrt{6+4\sqrt2}\right)}=$
$= \frac{2\left(\sqrt2-\sqrt{6+4\sqrt2}\right)}{2-\left(6+4\sqrt2\right)}=$
$= \frac{2\left(\sqrt2-\sqrt{6+4\sqrt2}\right)}{2-6-4\sqrt2}=$
$= \frac{2\left(\sqrt2-\sqrt{6+4\sqrt2}\right)}{-4-4\sqrt2}=$
$= \frac{2\left(\sqrt2-\sqrt{(2+\sqrt2)^2}\right)}{-4\left(1+\sqrt2\right)}=$
$= \frac{2\left(\sqrt2-\left|2+\sqrt2\right|\right)}{-4\left(1+\sqrt2\right)}= \frac{\sqrt2-2-\sqrt2}{-2\left(1+\sqrt2\right)}=$
$= \frac{-2}{-2\left(1+\sqrt2\right)}=\frac{1}{1+\sqrt2}=$
$= \frac{1-\sqrt2}{\left(1+\sqrt2\right)\left(1-\sqrt2\right)}=$
$= \frac{1-\sqrt2}{1-2}=\frac{1-\sqrt2}{-1}=\sqrt2-1;$
4. $\frac{2+\sqrt3}{\sqrt6-\sqrt3+\sqrt2-1}=$
$= \frac{2+\sqrt3}{(\sqrt{2\cdot3}-\sqrt3)+(\sqrt2-1)}=$
$= \frac{2+\sqrt3}{\sqrt3(\sqrt2-1)+(\sqrt2-1)}=$
$= \frac{2+\sqrt3}{(\sqrt2-1)(\sqrt3+1)}=$
$= \frac{(2+\sqrt3)(\sqrt2+1)}{(\sqrt2-1)(\sqrt2+1)(\sqrt3+1)}=$
$= \frac{(2+\sqrt3)(\sqrt2+1)(\sqrt3-1)}{2}=$
$= \frac{(2+\sqrt3)(\sqrt3-1)(\sqrt2+1)}{2}=$
$= \frac{(2\sqrt3-2+3-\sqrt3)(\sqrt2+1)}{2}=$
$= \frac{(\sqrt3+1)(\sqrt2+1)}{2}.$