Answer
$\dfrac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}=-\dfrac{1}{2x-2\sqrt{x^{2}+x}+1}$
Work Step by Step
$\dfrac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}$
Multiply the numerator and the denominator of the given expression by $\sqrt{x}-\sqrt{x+1}$, which is the conjugate of the numerator:
$\dfrac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}\cdot\dfrac{\sqrt{x}-\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}=...$
Evaluate the products and simplify:
$...=\dfrac{(\sqrt{x})^{2}-(\sqrt{x+1})^{2}}{(\sqrt{x}-\sqrt{x+1})^{2}}=...$
$...=\dfrac{x-x-1}{(\sqrt{x})^{2}-2(\sqrt{x})(\sqrt{x+1})+(\sqrt{x+1})^{2}}=...$
$...=-\dfrac{1}{x-2\sqrt{x^{2}+x}+x+1}=-\dfrac{1}{2x-2\sqrt{x^{2}+x}+1}$
