Answer
$\displaystyle \frac{2x+h-2\sqrt{x^2+xh}}{h}$
Work Step by Step
We rationalize the denominator:
$\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=\frac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}-\sqrt{x})}{(\sqrt{x+h}+\sqrt{x})(\sqrt{x+h}-\sqrt{x})}=\frac{(\sqrt{x+h})^2-2\sqrt{x}\sqrt{x+h}+(\sqrt{x})^2}{(x+h)-x}=\frac{x+h+x-2\sqrt{x}\sqrt{x+h}}{h}=\frac{2x+h-2\sqrt{x}\sqrt{x+h}}{h}=\frac{2x+h-2\sqrt{x(x+h)}}{h}=\frac{2x+h-2\sqrt{x^2+xh}}{h}$
