Question

rationalise the denominator ​

Answer 1

Answer:

$\frac{a-\sqrt{a^2-x^2} }{x}$

Step-by-step explanation:

$\frac{\sqrt{a+x} - \sqrt{a-x}}{\sqrt{a+x} + \sqrt{a-x}}$

= $\frac{\sqrt{a+x} - \sqrt{a-x}}{\sqrt{a+x} + \sqrt{a-x}} * \frac{\sqrt{a+x} - \sqrt{a-x}}{\sqrt{a+x} - \sqrt{a-x}}$

= ${ \frac{(\sqrt{a+x} - \sqrt{a-x})) ^2} {(\sqrt{a+x} + \sqrt{a-x}) (\sqrt{a+x} - \sqrt{a-x})} }$

= ${ \frac{(\sqrt{a+x})^2 - 2 ( \sqrt{a+x}) ( \sqrt{a-x}) + ( \sqrt{a-x})^2 } {(\sqrt{a+x})^2 - ( \sqrt{a-x})^2 } }$

= ${ \frac{ {a+x} - 2 ( \sqrt{(a+x)(a-x)} + {a-x} }{ (a+x) - ( {a-x}) } }$

= ${ \frac{ {2a} - 2 ( \sqrt{(a^2-x^2)} }{ 2x } }$

= $\frac{2 (a-\sqrt{a^2-x^2}) }{2x}$

= $\frac{a-\sqrt{a^2-x^2} }{x}$    , here the denominator is rationalised