Answer
$ \dfrac{2x+h-2\sqrt{x^2+xh}}{h}$
Work Step by Step
Rationalize the denominator by multiplying $\sqrt{x+h}-\sqrt x$ to both the numerator and the denominator:
$\dfrac{\sqrt{x+h}-\sqrt x}{\sqrt{x+h}+\sqrt x}\\
\\=\dfrac{\sqrt{x+h}-\sqrt x}{\sqrt{x+h}+\sqrt x}\cdot \dfrac{\sqrt{x+h}-\sqrt x}{\sqrt{x+h}-\sqrt x}\\
=\dfrac{\left(\sqrt{x+h}-\sqrt x\right)^2}{\left(\sqrt{x+h}+\sqrt x\right)\left(\sqrt{x+h}-\sqrt x\right)}$
Use special formula $(a-b)^2=a^2-2ab+b^2$ in the numerator and $(a+b)(a-b)=a^2-b^2$ in the denominator.
$=\dfrac{\left(\sqrt{x+h}\right)^2-2(\sqrt{x+h})(\sqrt x)+\left(\sqrt x\right)^2}{\left(\sqrt{x+h}\right)^2-\left(\sqrt x\right)^2}$
$=\dfrac{x+h-2(\sqrt{x+h})(\sqrt x)+x}{x+h-x}$
$=\dfrac{2x+h-2(\sqrt{x+h})(\sqrt x)}{h}$
Use the rule $\sqrt{a} \cdot \sqrt{b}=\sqrt{a\cdot b}$ then simplify:
$=\dfrac{2x+h-2\sqrt{x(x+h)}}{h}$
$=\dfrac{2x+h-2\sqrt{x^2+xh}}{h}$
Hence, the correct answer is $ \frac{2x+h-2\sqrt{x^2+xh}}{h}$.
