Question

plsss tell the ans. dont time pass. best one will be marked branliest

Answer 1

Answer :

N equals 1 [ option (A) ]

Step-by-step explanation :

$\sf N=\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}$

Let

$\sf \dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}=a \\\\\\ \sf \sqrt{3-2\sqrt{2}} =b$

First, find the value of a.

 Rationalizing factor = $\sqrt{\sqrt{5}-1}}$

 $\sf a=\dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}} \times \dfrac{\sqrt{\sqrt{5}-1}}{\sqrt{\sqrt{5}-1}} \\\\\\ \sf a=\dfrac{(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})(\sqrt{\sqrt{5}-1})}{(\sqrt{\sqrt{5}+1})(\sqrt{\sqrt{5}-1})} \\\\\\ \sf a=\dfrac{(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})(\sqrt{\sqrt{5}-1})}{\sqrt{(\sqrt{5}+1)(\sqrt{5}-1)}} \\\\\\ \sf a=\dfrac{(\sqrt{\sqrt{5}+2})(\sqrt{\sqrt{5}-1})+(\sqrt{\sqrt{5}-2})(\sqrt{\sqrt{5}-1})}{\sqrt{\sqrt{5}(\sqrt{5}-1)+1(\sqrt{5}-1)}}$

 $\sf a=\dfrac{\sqrt{(\sqrt{5}+2)(\sqrt{5}-1)}+\sqrt{(\sqrt{5}-2)(\sqrt{5}-1})}{\sqrt{\sqrt{5}^2-\sqrt{5}+\sqrt{5}-1}} \\\\\\ \sf a=\dfrac{\sqrt{(\sqrt{5}+2)(\sqrt{5}-1)}+\sqrt{(\sqrt{5}-2)(\sqrt{5}-1})}{\sqrt{5-1}} \\\\\\ \sf a=\dfrac{\sqrt{(\sqrt{5}+2)(\sqrt{5}-1)}+\sqrt{(\sqrt{5}-2)(\sqrt{5}-1})}{\sqrt{4}} \\\\\\ \sf a=\dfrac{\sqrt{(\sqrt{5}+2)(\sqrt{5}-1)}+\sqrt{(\sqrt{5}-2)(\sqrt{5}-1})}{2}$

 $\sf a=\dfrac{\sqrt{\sqrt{5}(\sqrt{5}-1)+2(\sqrt{5}-1)}+\sqrt{\sqrt{5}(\sqrt{5}-1)-2(\sqrt{5}-1)}}{2} \\\\\\ \sf a=\dfrac{\sqrt{\sqrt{5}^2-\sqrt{5}+2\sqrt{5}-2}+\sqrt{\sqrt{5}^2-\sqrt{5}-2\sqrt{5}+2}}{2} \\\\\\ \sf a=\dfrac{\sqrt{5+\sqrt{5}-2}+\sqrt{5-3\sqrt{5}+2}}{2} \\\\\\ \sf a=\dfrac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}$

 $\sf multiply \ and \ divide \ the \ numerator \ terms \ by \ 2, \\\\ \sf a=\dfrac{\bigg(\sqrt{(3+\sqrt{5}) \times \dfrac{2}{2}}\bigg)+\bigg(\sqrt{(7-3\sqrt{5}) \times \dfrac{2}{2}}\bigg)}{2} \\\\\\ \sf a=\dfrac{\dfrac{1}{\sqrt{2}} \bigg(\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}\bigg)}{2} \\\\\\ \sf a=\dfrac{ \bigg(\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}\bigg)}{2\sqrt{2}}$

 Modify the numerator terms to use any algebraic identity.

  $\sf a=\dfrac{\sqrt{\sqrt{5}^2+1^2+2(1)(\sqrt{5})}+\sqrt{\sqrt{5}^2+3^2-2(\sqrt{5})(3)}}{2\sqrt{2}} \\\\\\ \sf a=\dfrac{\sqrt{(1+\sqrt{5})^2}+\sqrt{(3-\sqrt{5})^2}}{2\sqrt{2}}\\\\ \sf a=\dfrac{1+\sqrt{5}+3-\sqrt{5}}{2\sqrt{2}}\\\\ \sf a=\dfrac{4}{2\sqrt{2}}\\\\ \sf a=\dfrac{2}{\sqrt{2}} \\\\ \sf a=\dfrac{2}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} \\\\ \sf a=\dfrac{2\sqrt{2}}{2} \\\\ \sf a=\sqrt{2}$

Now, find the value of b.

$\sf b=\sqrt{3-2\sqrt{2}} \\\\ \sf b=\sqrt{1^2+\sqrt{2}^2-2(1)(\sqrt{2})} \\\\ \sf b=\sqrt{(\sqrt{2}-1)^2} \\\\ \sf b=\sqrt{2}-1$

N = a - b

N = √2 - (√2 - 1)

N = √2 - √2 + 1

N = 1

∴ The value of N is 1

___________________

IDENTITES USED :

x² + y² + 2xy = (x + y)²

x² + y² - 2xy = (x - y)²