Question

Find the Sum of the reciprocal of x+3/x²+1 and x+2/x²-3 .​

Answer 1

Answer:

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Reciprocal of

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

Reciprocal of

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

So,

Their Sum will be:

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

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

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

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

Answer 2

Answer:

Let the number be x

Reciprocal = 1 / x

So sum = x + ( 1 / x )

= ( x² + 1 ) / x

Difference = ( x² - 1 ) / x

According to ques,

( x² + 1 ) / x = 3 ( x² - 1 ) / x

Removing 1 / x form both sides,

x² + 1 = 3x² - 3

3 + 1 = 3x² - x²

4 / 2 = x²

√2 = x

So the no. is √2 or 1.41

hope help u ✌