Question

anyone Pls solve this

Answer 1

$= > \frac{4 + \sqrt{2} }{2 + \sqrt{2} } = n - \sqrt{b} \\ \\ = > \frac{4 + \sqrt{2} }{2 + \sqrt{2} } \times \frac{2 - \sqrt{2} }{2 - \sqrt{2} } = n - \sqrt{b} \\ \\ = > \frac{8 - 4 \sqrt{2} + 2 \sqrt{2} - 2}{4 - 2} = n - \sqrt{b} \\ \\ = > \frac{6 - 2 \sqrt{2} }{2} = n - \sqrt{b} \\ \\ = > \frac{2(3 - \sqrt{2}) }{2} = n - \sqrt{b} \\ \\ = > 3 - \sqrt{2} = n - \sqrt{b} \\ \\ = > n = 3 \: \: and \: b \: = 2$

Answer 2

hope this helps you.......