Simplifying rational expressions 2

**Lil** · Oct 10th 2010, 10:20 AM

How about this? $\frac{a-b}{a+b+2a^{\frac{1}{2}}b^{\frac{1}{2}}}:\frac{a^{-\frac{1}{2}}-b^{-\frac{1}{2}}}{a^{-\frac{1}{2}}+b^{-\frac{1}{2}}}$

**harish21** · Oct 10th 2010, 10:53 AM

First of all,Are you taking ratio of those two terms?

Secondly, Show your work!

Also,

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

and

$\displaystyle{a^{-\frac{1}{2}} = \frac{1}{\sqrt{a}}}$

**Lil** · Oct 10th 2010, 01:14 PM

$...=\frac{(a-b)*(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})^2*(\sq rt{a}-\sqrt{b})}=\frac{(a-b)}{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}=\frac{a-b}{a-b}=1$
But answer have to be -1. What's wrong?

**harish21** · Oct 10th 2010, 01:44 PM

Originally Posted by Lil

$...=\frac{(a-b)*(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})^2*(\sq rt{a}-\sqrt{b})}=\frac{(a-b)}{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}=\frac{a-b}{a-b}=1$
But answer have to be -1. What's wrong?

$\displaystyle{\frac{a^{-\frac{1}{2}}-b^{-\frac{1}{2}}}{a^{-\frac{1}{2}}+b^{-\frac{1}{2}}} =\frac{\sqrt{b}-\sqrt{a}}{\sqrt{b}+\sqrt{a}}}$

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