Question

$if \: x = 2 + \sqrt{3} and \: xy = 1 \\ then \: find \: the \: value \: of \: \frac{ x }{ \sqrt{2} + \sqrt{x} } + \frac{y}{ \sqrt{2} + \sqrt{y} } \\ \\ \: no \: spam \: plzz$
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Answer 1

Probable Question :

$If \: x = 2 + \sqrt{3} \:and \: xy = 1 \: then \: find \: the \\ value \: of \: \frac{ x }{ \sqrt{2} + \sqrt{x} } + \frac{y}{ \sqrt{2} - \sqrt{y} }$

Answer :

√2

(Also , refer to the attachments for the solution of given question)

Solution :

• $Given : \: x = 2 + \sqrt{3} , \: xy = 1$

• $To find : \: \frac{ x }{ \sqrt{2} + \sqrt{x} } + \frac{y}{ \sqrt{2} - \sqrt{y} }$

We have ;

=> xy = 1

=> y = 1/x

=> y = 1/(2 + √3)

=> y = (2 - √3)/(2 + √3)(2 - √3)

=> y = (2 - √3)/[2² - (√3)²]

=> y = (2 - √3)/(4 - 3)

=> y = (2 - √3)/1

=> y = 2 - √3

Now,

=> (√x - √y)² = (√x)² - 2•√x•√y + (√y)²

=> (√x - √y)² = x - 2√(xy) + y

=> (√x - √y)² = 2 + √3 - 2√1 + 2 - √3

=> (√x - √y)² = 2 - 2 + 2

=> (√x - √y)² = 2

=> √x - √y = √2

( Note : While manipulating any formula or identity , always take positive value unless it is provided in question. )

Now ;

$\frac{x}{ \sqrt{2} + \sqrt{x} } + \frac{y}{ \sqrt{2} - \sqrt{y} } \\ = \frac{x( \sqrt{2} - \sqrt{y} ) + y( \sqrt{2} + \sqrt{x} )}{( \sqrt{2} + \sqrt{x} )( \sqrt{2} - \sqrt{y}) } \\ = \frac{x \sqrt{2} - x \sqrt{y} + y \sqrt{2} + y \sqrt{x} }{ \sqrt{2} \sqrt{2} - \sqrt{2} \sqrt{y} + \sqrt{2} \sqrt{x} - \sqrt{x} \sqrt{y} } \\ = \frac{x \sqrt{2} + y \sqrt{2} - x \sqrt{y} + y \sqrt{x} }{2 + \sqrt{2} \sqrt{x} - \sqrt{2} \sqrt{y} - \sqrt{x} \sqrt{y} } \\ = \frac{ \sqrt{2}(x + y) - \sqrt{xy} ( \sqrt{x} - \sqrt{y} )}{2 + \sqrt{2}( \sqrt{x} - \sqrt{y} ) - \sqrt{xy} } \\ = \frac{ \sqrt{2} \times 4 - \sqrt{1} \times \sqrt{2} }{2 + \sqrt{2} \times \sqrt{2} - 1 } \\ = \frac{4 \sqrt{2} - \sqrt{2} }{2 + 2 - 1} \\ = \frac{3 \sqrt{2} }{3} \\ = \sqrt{2}$

Hence ,

The required answer is √2 .