Question

the Question is in image please somebody answer ​

Answer 1

Step-by-step explanation:

6.

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

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

Let's take x +1/x be y

${y}^{2} - 2y = > y(y - 2) = >$

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

7.

$y - \frac{1}{y} = 0 = > {(y - \frac{1}{y}) }^{2} = 0 = > {y}^{2} +$

$\frac{1}{ {y}^{2} } - 2 \times y \times \frac{1}{y} = 0 = > {y}^{2} + \frac{1}{ {y}^{2} } = 2$

On squaring on both sides

$( {y}^{2} + \frac{1}{ {y}^{2} })^{2} = 4 = > {y}^{4} + \frac{1}{ {y}^{4}} + 2 = 4$

$= > {y}^{4} + \frac{1}{ {y}^{4}} = 2$

Or

$y - \frac{1}{y} = 0 = > y = \frac{1}{y} = > {y}^{2} = 1 = >$

$y = 1 \: or \: - 1 = > {y}^{2} + \frac{1}{ {y}^{2}} = 2$

${y}^{4} + \frac{1}{ {y}^{4} } = 2$

Hope it helps you.