Question

Освободитесь от иррациональности в знаменателей дробей:

Answer 1

$1)\; \; \frac{\sqrt{2+\sqrt3}}{\sqrt{2-\sqrt3}}=\sqrt{\frac{(2+\sqrt3)^2}{(2-\sqrt3)(2+\sqrt3)}}=\sqrt{\frac{(2+\sqrt3)^2}{4-3}}=\sqrt{\frac{(2+\sqrt3)^2}{1}}=2+\sqrt3$

$2)\; \; \frac{2\sqrt{21}}{\sqrt7-2\sqrt3+1}=\frac{2\sqrt{21}\cdot (\sqrt7+1+2\sqrt3)}{(\sqrt7+1-2\sqrt3)(\sqrt7+1+2\sqrt3)}=\frac{2\sqrt{21}\cdot (\sqrt7+1+2\sqrt3)}{(\sqrt7+1)^2-12}=\\\\=\frac{2\sqrt{21}\cdot (\sqrt7+1+2\sqrt3)}{7+1+2\sqrt7-12}=\frac{2\sqrt{21}\cdot (\sqrt7+1+2\sqrt3)}{2\cdot (\sqrt7-2)}=\frac{\sqrt{21}\cdot (\sqrt7+1+2\sqrt3)\cdot (\sqrt7+2)}{(\sqrt7-2)(\sqrt7+2)}=\\\\=\frac{\sqrt{21}\cdot (\sqrt7+1+2\sqrt3)(\sqrt7+2)}{7-4}=\frac{\sqrt{21}\cdot (\sqrt7+1+2\sqrt3)\cdot (\sqrt7+2)}{3}$

$3)\; \; \frac{2x+\sqrt{x^2-1}}{\sqrt{(x+1)^3}-\sqrt{(x-1)^3}}=\frac{2x+\sqrt{x^2-1}}{(\sqrt{x+1}-\sqrt{x-1})(x+1+\sqrt{(x+1)(x-1)}+x-1)}=\\\\=\frac{2x+\sqrt{x^2-1}}{(\sqrt{x+1}-\sqrt{x-1})\cdot (2x+\sqrt{x^2-1})}=\frac{\sqrt{x+1}+\sqrt{x-1}}{(\sqrt{x+1}-\sqrt{x-1})(\sqrt{x+1}+\sqrt{x-1})}=\\\\=\frac{\sqrt{x+1}+\sqrt{x-1}}{(x+1)-(x-1)}=\frac{\sqrt{x+1}+\sqrt{x-1}}{2}$