Question

x minus one by X and and x square minus one by x square if X + 1 by x is equal to root 5​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

AS WE HAVE TO REMOVE THE UNDER ROOT FIRST

SQUARUNG BOTH THE SIDES

$(x + \frac{1}{x} ) {}^{2} = ( \sqrt{5} ) {}^{2}$

${x}^{2} + \frac{1}{ {x}^{2} } + 2 \times x \times \frac{1}{x} = 5$

${x}^{2} + \frac{1}{ {x}^{2} } + 2 = 5$

${x}^{2} + \frac{1}{ {x}^{2} } = 5 - 2$

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

NOW WE WILL FIND THE VALUE OF X-1/X=???

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

$(x - \frac{1}{x} ) { }^{2} = 3 - 2$

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

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

$x - \frac{1}{x} = \binom{ + }{ - } 1$

NOW WE WILL FIND THE VALUE OF X²-1/X²=?????

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

$(\binom{ + }{ - } 1) \times ( \sqrt{5} )$

$( \binom{ + }{ - } \sqrt{5} )$

${x}^{2} - \frac{1}{ {x}^{2} } = \binom{ + }{ - } \sqrt{5} \: \: answer$

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