Question

Answer the question correctly plsssss... ​

Answer 1

$x = 1 - \sqrt{2}$

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

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

${x}^{2} =3 - 2 \sqrt{2}$

$\frac{1}{ {x}^{2} } = \frac{1}{3 - 2 \sqrt{2} } \times \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} }$

$= > \frac{3 + 2 \sqrt{2} }{9 - 8}$

$= > \frac{1}{ {x}^{2} } = 3 + 2 \sqrt{2}$

Now,

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

$= > 3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} - 2$

$= > 3 + 3 - 2$

$= > 4$

Thus,the value is 4.

Answer 2

Answer:

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