Answer
$\dfrac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}=-2x-2\sqrt{x^{2}+x}-1$
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 denominator:
$\dfrac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}-\sqrt{x+1}}\cdot\dfrac{\sqrt{x}+\sqrt{x+1}}{\sqrt{x}+\sqrt{x+1}}=...$
Evaluate the products:
$...=\dfrac{(\sqrt{x}+\sqrt{x+1})^{2}}{(\sqrt{x})^{2}-(\sqrt{x+1})^{2}}=...$
$...=\dfrac{(\sqrt{x})^{2}+2(\sqrt{x})(\sqrt{x+1})+(\sqrt{x+1})^{2}}{x-x-1}=...$
$...=\dfrac{x+2\sqrt{x^{2}+x}+x+1}{-1}=-2x-2\sqrt{x^{2}+x}-1$
