Question

please solve this question​

Answer 1

Answer:

N = 1

Step-by-step explanation:

Given :

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

Then simpliy N,.

Solution :

⇒ $N = \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}} - \sqrt{3-2\sqrt{2}}$

⇒ $[(\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}})^2]^{\frac{1}{2} - \sqrt{3-2\sqrt{2}}$

⇒ $\sqrt{(\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}})^2} - \sqrt{3-2\sqrt{2}}$

⇒ $\sqrt{\frac{(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})^2}{(\sqrt{\sqrt{5}+1})^2}} - \sqrt{3-2\sqrt{2}}$

⇒ $\sqrt{\frac{(\sqrt{\sqrt{5}+2})^2+(\sqrt{\sqrt{5}-2})^2 + 2(\sqrt{\sqrt{5}+2})(\sqrt{\sqrt{5}-2}) }{(\sqrt{5}+1)}} - \sqrt{3-2\sqrt{2}}$

⇒ $\sqrt{\frac{(\sqrt{5}+2)+(\sqrt{5}-2) + 2(\sqrt{\sqrt{5}+2)(\sqrt{5}-2}) }{(\sqrt{5}+1)}} - \sqrt{3-2\sqrt{2}}$

⇒ $\sqrt{\frac{2\sqrt{5}+ 2(\sqrt{(\sqrt{5})^2-2^2)} }{(\sqrt{5}+1)}} - \sqrt{3-2\sqrt{2}}$

⇒ $\sqrt{\frac{2\sqrt{5}+ 2(\sqrt{(5-4)} }{(\sqrt{5}+1)}} - \sqrt{2+1-2\sqrt{2}}$

⇒ $\sqrt{\frac{2\sqrt{5}+ 2(\sqrt{1}) }{(\sqrt{5}+1)}} - \sqrt{(\sqrt{2})^2+(1)^2-2(\sqrt{2})(1)}$

⇒ $\sqrt{\frac{2(\sqrt{5}+1) }{(\sqrt{5}+1)}} - \sqrt{(\sqrt{2}-1)^2}$

⇒ $\sqrt{2} - (\sqrt{2}-1)$

⇒ $\sqrt{2} - \sqrt{2} + 1 = 1$