МАТЕМАТИКА 9 КЛАСС (2.В.36 а и б)

Решите задачу:

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

$b)\; \; \frac{\frac{1}{\sqrt{a-1}}+\sqrt{a-1}}{\frac{1}{\sqrt{a+1}}-\frac{1}{\sqrt{a-1}}}:\frac{\sqrt{a-1}}{(a-1)\sqrt{a+1}-(a+1)\sqrt{a-1}}=\\\\=\frac{(1+(a-1))\sqrt{(a-1)(a+1)}}{\sqrt{a-1}(\sqrt{a-1}-\sqrt{a+1})}\cdot \frac{\sqrt{(a-1)(a+1)}(\sqrt{a-1}-\sqrt{a+1})}{\sqrt{a-1}}=\\\\=\frac{a\cdot (\sqrt{(a-1)(a+1)})^2}{(\sqrt{a-1})^2}=\frac{a\cdot (a-1)(a+1)}{a-1}=a(a+1)$