Question

ifx=2+root 3,xy=1 then x/root2+rootx+y/root2-rooty

Answer 1

Find the value of square roots of x and y.   Then use rationalization of denominator to find the value of y.

x = 2 + √3  
xy = 1

y = 1 /(2 + √3) = (2 - √3)/[(2 - √3)(2 + √3) 
   = 2 - √3

Let   √x = √a + √b
then   x = 2 + √3 = a + b + 2 √ab
        so   a + b = 2    and   ab = 3 /4
               (a - b)² = 4 - 3 = 1
                a - b = 1           =>  a = 3/2      b = 1/2
      so √x = (√3 +1)/√2
      √y = 1/√x = (√3 - 1)/√2

Now substitute in the given expression:

$\frac{x}{\sqrt2+\sqrt{x}}=\frac{(2+\sqrt3)*\sqrt2}{\sqrt2*\sqrt2+\sqrt3+1}=\frac{\sqrt2(2+\sqrt3)}{3+\sqrt3}\\\\=\frac{\sqrt2(2+\sqrt3)}{3+\sqrt3}*\frac{(3-\sqrt3)}{3-\sqrt3}=\frac{\sqrt2(3+\sqrt3)}{6}\\\\\frac{y}{\sqrt2-\sqrt{y}}=\frac{(2-\sqrt3)\sqrt2}{\sqrt2*\sqrt2-\sqrt3+1}=\frac{\sqrt2(2-\sqrt3)}{3-\sqrt3}\\\\=\frac{\sqrt2(2-\sqrt3)}{3-\sqrt3}*\frac{3+\sqrt3}{3+\sqrt3}=\frac{\sqrt2*(3-\sqrt3}{6}\\\\Adding\ them\ get\ Answer=\sqrt2$

Answer 2

Answer:

√2

Step-by-step explanation:

Given :

• x = 2 + √3
• xy = 1

⇒ xy = 1

⇒ y = 1 / x

⇒ y = 1 / ( 2 + √3 )

Rationalising the denominator

⇒ 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

Using the identity given below

⇒ ( √x - √y )² = ( √x )² + ( √y )² - 2( √x )( √y )

⇒ ( √x - √y )² = x + y - 2√xy

⇒ ( √x - √y )² = 2 + √3 + 2 - √3 - 2√1

⇒ ( √x - √y )² = 4 - 2( 1 )

⇒ ( √x - √y )² = 4 - 2

⇒ ( √x - √y )² = 2

⇒ √x - √y = √2

Now let's find out the value of expression.

$\Rightarrow \sf \dfrac{x}{ \sqrt{2} + \sqrt{x} } + \dfrac{y}{ \sqrt{2} - \sqrt{y} }$

$\Rightarrow \sf \dfrac{x( \sqrt{2} - \sqrt{y} ) + y( \sqrt{2} + \sqrt{x} )}{( \sqrt{2} + \sqrt{x})(\sqrt{2} - \sqrt{y} )}$

$\Rightarrow \sf \dfrac{x\sqrt{2} - x\sqrt{y} + y\sqrt{2} + y\sqrt{x} }{ \sqrt{2} (\sqrt{2} - \sqrt{y} ) + \sqrt{x}( \sqrt{2} - \sqrt{y}) }$

$\Rightarrow \sf \dfrac{\sqrt{2} (x + y)- \sqrt{xy} ( \sqrt{x} - \sqrt{y} )}{2- \sqrt{2} \sqrt{y} + \sqrt{x} \sqrt{2} - \sqrt{xy} }$

$\Rightarrow \sf \dfrac{4\sqrt{2} - \sqrt{2} }{2 + \sqrt{2} ( \sqrt{2}) - 1}$

$\Rightarrow \sf \dfrac{3\sqrt{2} }{2 + 2 - 1}$

$\Rightarrow \sf \dfrac{3\sqrt{2} }{3}$

$\Rightarrow \sf \sqrt{2}$

Hence the value of the expression is √2.