Question

Освободитесь от иррациональности в знаменателя дроби

Answer 1

a)

$\frac{2}{\sqrt 7}=\frac{2\sqrt 7}{7}$

б)

$\frac{1}{\sqrt 5 - \sqrt 3}=\frac{\sqrt 5 + \sqrt 3}{5-3} =\frac{\sqrt 5 + \sqrt 3}{2}$

в)

$\frac{1}{\sqrt{4-2\sqrt 3}}=\frac{\sqrt{4+2\sqrt 3}}{\sqrt{16-12}}= \frac{\sqrt{4+2\sqrt 3}}{\sqrt{4}}=\frac{\sqrt{4+2\sqrt 3}}{2}$

Объяснение:

Answer 2

$a)\;\dfrac{2^{|\underline{\sqrt7}}}{\sqrt7}=\dfrac{2\sqrt7}{\sqrt7\cdot\sqrt7}=\dfrac{2\sqrt7}{\sqrt{49}}=\dfrac{2\sqrt7}7$

$b)\;\dfrac{1^{|\underline{(\sqrt5+\sqrt3)}}}{\sqrt5-\sqrt3}=\dfrac{\sqrt5+\sqrt3}{(\sqrt5-\sqrt3)(\sqrt5+\sqrt3)}=\dfrac{\sqrt5+\sqrt3}{\sqrt5^2-\sqrt3^2}=\dfrac{\sqrt5-\sqrt3}{5-3}=\dfrac{\sqrt5+\sqrt3}{2}$

$c)\;\dfrac1{\sqrt{4-2\sqrt3}}=\dfrac{1}{\sqrt{4-\sqrt{12}}}=\dfrac{1^{|\underline{\sqrt{4+\sqrt{12}}}}}{(\sqrt{4-\sqrt{12}})(\sqrt{4+\sqrt{12}})}=\dfrac{\sqrt{4+\sqrt{12}}}{\sqrt{16-12}}=$

$=\dfrac{\sqrt{4+2\sqrt3}}{\sqrt{4}}=\dfrac{\sqrt{4+2\sqrt3}}{2}$