Question

(x2+1/x2)-3(x-1/x)-2=0​

Answer 1

$( {x}^{2} + \frac{ {1}^{} }{ {x}^{2} }) - 3(x - \frac{1}{x} ) - 2 = 0 \\ \\ = > {( {x} - \frac{1}{x}) }^{2} + 2x \times \frac{1}{x} - 3(x - \frac{1}{x} ) - 2 = 0 \\ \\ = > {(x - \frac{1}{x} )}^{2} + 2 - 3(x - \frac{1}{x} ) - 2 = 0 \\ \\ = > {(x - \frac{1}{x} )}^{2} -3(x - \frac{1}{x} ) = 0$

$let \: x - \frac{1}{x} = t \\ \\ = > {t}^{2} - 3t = 0 \\ \\ = > t(t - 3) = 0$

Now, t = 0

$x - \frac{1}{x} = 0 \\ \\ = > {x}^{2} - 1 = 0 \\ \\ = > x = \binom{ + }{ - } \sqrt{1} = \binom{ + }{ - } 1$

And , t = 3

$= > x - \frac{1}{x} = 3 \\ \\ = > {x }^{2} - 1 = 3x \\ \\ = > {x}^{2} - 3x - 1 = 0 \\ \\ x = \frac{ 3 \binom{ + }{ - } \sqrt{9 + 4} }{2} = \frac{3 \binom{ + }{ - } \sqrt{13} }{2}$

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