Question

rationalize the denominator : ​

Answer 1

Answer:

$x-\sqrt{x^2-1}$

Step-by-step explanation:

$\dfrac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\\\\=\dfrac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x-1}}\times\dfrac{\sqrt{x+1}-\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}$

$=\dfrac{(\sqrt{x+1}-\sqrt{x-1})^2}{(\sqrt{x+1})^2-(\sqrt{x-1})^2}$

$=\dfrac{x+1-2\sqrt{(x+1)(x-1)}+x-1}{x+1-(x-1)}\\\\=\dfrac{2x-2\sqrt{x^2-1}}{2}\\\\=x-\sqrt{x^2-1}$