Simplifying $\frac{a^x}{a^x+\sqrt{a}} + \frac{a^{1-x}}{a^{1-x}+\sqrt{a}}$ While solving a problem related to functions, I came across the following algebraic expression: 
$$\frac{a^x}{a^x+\sqrt{a}} + \frac{a^{1-x}}{a^{1-x}+\sqrt{a}}$$
Although I tried, I couldn't find an easy way to simplify it.

Can anyone simplify the stated equation in the easiest way so that anyone can understand it?

 A: It is $$\frac{a^x(a^{1-x}+\sqrt{a})+a^{1-x}(a^x+\sqrt{a})}{(a^x+\sqrt{a})(a^{1-x}+\sqrt{a})}$$
this simplifies to
$$\frac{2a+a^{x+1/2}+a^{3/2-x}}{(a^x+\sqrt{a})(a^{1-x}+\sqrt{a})}$$
A: Since$$\frac{a^{1-x}}{a^{1-x}+a^{1/2}}=\frac{a^{1-x}a^{x-1/2}}{a^{x-1/2}a^{1-x}+a^{x-1/2}a^{1/2}}=\frac{a^{1/2}}{a^{1/2}+a^x}$$(by multiplying the numerator and denominator by $a^{x-1/2}$), the sum is $\frac{a^x+a^{1/2}}{a^x+a^{1/2}}=1$. More generally,$$\frac{u}{u+v}+\frac{w}{w+v}-1=\frac{uw-v^2}{uw+uv+wv+v^2}$$vanishes if $uw=v^2$, as happens if $u=a^x,\,w=a^{1-x},\,v=\sqrt{a}$.
A: $\ {a ^ x \over a^x + \sqrt a} + {a ^ {1-x} \over 
    a ^ {1-x} + \sqrt a}$
Well, often it's a good idea to put over common denominator:
$\frac {a^x(a^{1-x} + \sqrt a) + a^{1-x}(a^x+\sqrt a)}{(a^x + \sqrt a)(a^{1-x} + \sqrt a)} =$
$\frac {(a^{x + 1-x} + a^x\sqrt a) + (a^{1-x+x} + a^{1-x}\sqrt a)}{a^{x+1-z} + a^x\sqrt a + a^{1-x}\sqrt a + a}=$
$\frac {2a^1 + (a^x + a^{1-x})\sqrt a}{2a+(a^x + a^{1-x})\sqrt a} = 1$.
A: The second fraction should be $$\frac{a^{1 \over 2}}{a^{1 \over 2}+a^{x}}$$ . 
