Question

please give me answer it is very urgent​

Answer 1

Answer:

Given,

$x = 1 - \sqrt{2}$

therefore,

$\frac{1}{x} = \frac{1}{1 - \sqrt{2} }$

$= > \frac{1}{x} = \frac{1(1 + \sqrt{2}) }{(1 - \sqrt{2})(1 + \sqrt{2} )}$

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

$= > \frac{1}{x} = \frac{1 + \sqrt{2} }{1 - 2}$

$= > \frac{1}{x} = \frac{1 + \sqrt{2} }{ - 1}$

$= > \frac{1}{x} = - (1 + \sqrt{2} )$

Now,

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

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

${(x - \frac{1}{x} )}^{2} = {(2)}^{2}$

${(x - \frac{1}{x} )}^{2} = 4$

HOPE IT WORKS OUT FOR YOU AND MARK ME BRAINLIEST AND THANKS MY ANSWERS

Answer 2

Answer:

$x = 1 - \sqrt{2} \\ {(x - \frac{1}{x} )} = {1 - \sqrt{2} + \frac{1}{1 - \sqrt{2} }}\\ = 1 - \sqrt{2} + \frac{1 + \sqrt{2} }{(1 - \sqrt{2} )(1 + \sqrt{2} )} \\ = 1 - \sqrt{2} + \frac{1 + \sqrt{2} }{1 - 2} \\ = 1 - \sqrt{2} - 1 - \sqrt{2} \\ = - 2 \sqrt{2} \\ so \: {{(x - \frac{1}{x} )} }^{2} = {( - 2 \sqrt{2} )}^{2} = 8$