Question

evaluate it plsss......​

Answer 1

SOLUTION

We cannot directly put the values of x because the denominator will become zero so we have to do something.First of all let us rationalize it,

$\frac{ \sqrt{1 + 2x} - \sqrt{1 - 2x} }{ \sqrt{1 + x} - \sqrt{1 - x} } \\ = \frac{ ( \sqrt{1 + 2x} - \sqrt{1 - 2x} )( \sqrt{1 + x} + \sqrt{1 - x}) }{ {( \sqrt{1 + x}) }^{2} - { (\sqrt{1 - x} })^{2} } \\ = \frac{ \sqrt{1 + x} + \sqrt{1 - x} }{ { (\sqrt{1 + x} )}^{2} - { (\sqrt{1 - x} })^{2} } = \frac{ {x}^{1} - ( - a)^{1} }{ {x}^{2} - {( - a)}^{2} } \\ = \frac{1}{2} ( - a ^{1 - 2} ) \\ = \frac{1}{2}( - a ^{ - 1}) \\ = \frac{1}{2} \times ( \frac{ - 1}{ \sqrt{1 - x} } ) \times ( \sqrt{1 + 2x} - \sqrt{1 - 2x} )$

Now, we can put the value of x ,

$\frac{1}{2} \times ( \frac{ - 1}{ \sqrt{1 - 0} } ) \times ( \sqrt{1 + 2 \times 0 } - \sqrt{1 - 2 \times 0} ) \\ = \frac{1}{2} \times - (\frac{ 1}{1}) \times 0 \\ = 0$

This is the answer hope it will help