Simplifying Radicals

Question

Simplify $$\frac{1-a^{1/2}}{1+a^{1/2}}-\frac{a^{1/2}+a^{-1/2}}{a-1},\quad a>0,\ a\ne 1$$

I can't figure out how to simplify this if a>0 and a doesn't equal 1. Can someone help me?

Answer 1

$$a^{1/2}=\sqrt a\implies \frac{1-\sqrt a}{1+\sqrt a}-\frac{\sqrt a+\frac1{\sqrt a}}{a-1}=\frac{1-\sqrt a}{1+\sqrt a}-\frac{a+1}{\sqrt a(a-1)}=$$

$$=\frac{(\sqrt a-a)(a-1)-(a+1)(\sqrt a+1)}{\sqrt a(a-1)(\sqrt a+1)}=\frac{\require{cancel}\cancel{a\sqrt a}-\sqrt a-a^2+\cancel a-\cancel{a\sqrt a}-\cancel a-\sqrt a-1}{\sqrt a(a-1)(\sqrt a+1)}$$

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

We've used here:

$$\sqrt a\sqrt a=a\;,\;\;\frac a{\sqrt a}=\sqrt a\;,\;\;\frac{a+\frac bc}d=\frac{ac+b}{cd}\;,\;\;a^{-1/2}=\frac1{a^{1/2}},\;\;\text{etc.}$$

Answer 2

$=\dfrac{1-\sqrt{a}}{1+\sqrt{a}}-\dfrac{\sqrt{a}+\dfrac{1}{\sqrt{a}}}{a-1}$

$=\dfrac{1-\sqrt{a}}{1+\sqrt{a}}-\dfrac{\dfrac{a+1}{\sqrt{a}}}{a-1}$

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

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

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

$=\dfrac{(\sqrt{a}-a)(a-1)-a+a\sqrt{a}+1+\sqrt{a}}{(\sqrt{a}+a)(a-1)}\;\;\;\;\;(\because \sqrt{a}\times\sqrt{a}=a)$

$=\dfrac{a\sqrt{a}-\sqrt{a}-a^2+a-a+a\sqrt{a}+1+\sqrt{a}}{a\sqrt{a}-\sqrt{a}+a^2-a}$

$=\dfrac{2a\sqrt{a}-a^2+1}{\sqrt{a}(a-1)+a(a-1)}\;\;\;\;\;\Big(\because a\sqrt{a}-\sqrt{a}=\sqrt{a}(a-1)\;and\;a^2-a=a(a-1)\Big)$

$=\dfrac{2a\sqrt{a}-a^2+1}{(a-1)(\sqrt{a}+a)}\;\;\;\;\;\Big(\because \sqrt{a}(a-1)+a(a-1)=(a-1)(\sqrt{a}+a)\Big)$

Hope, it will help.