Question

Упростите:
$(\frac{1}{\sqrt{a}+\sqrt{a+1} }+\frac{1}{\sqrt{a}-\sqrt{a-1} } ):(1+\sqrt{\frac{a+1}{a-1} } )$

Answer 1

$1)\frac{1}{\sqrt{a}+\sqrt{a+1}}+\frac{1}{\sqrt{a}-\sqrt{a-1}}=\frac{\sqrt{a}-\sqrt{a+1}}{(\sqrt{a}+\sqrt{a+1})(\sqrt{a}-\sqrt{a+1})}+\frac{\sqrt{a}+\sqrt{a-1}}{(\sqrt{a}-\sqrt{a-1})(\sqrt{a}+\sqrt{a-1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}+\frac{\sqrt{a}+\sqrt{a-1}}{a-a+1}=\sqrt{a}+\sqrt{a-1}-\sqrt{a}+\sqrt{a+1}=\sqrt{a-1}+\sqrt{a+1}$

$2)1+\sqrt{\frac{a+1}{a-1} }=1+\frac{\sqrt{a+1} }{\sqrt{a-1} }=\frac{\sqrt{a-1}+\sqrt{a+1}}{\sqrt{a-1} }\\\\3)(\sqrt{a-1}+\sqrt{a+1}):\frac{\sqrt{a-1}+\sqrt{a+1}}{\sqrt{a-1} }=(\sqrt{a-1}+\sqrt{a+1})*\frac{\sqrt{a-1} }{\sqrt{a-1}+\sqrt{a+1}}=\sqrt{a-1}$