Question

root 3 + 1 by 2 root 2 minus root 3​

Answer 1

Answer:

$\bold{\dfrac{2\sqrt{2} + \sqrt{3} + 2\sqrt{6} + 3}{5}}$

Step-by-step explanation:

$\dfrac{\sqrt{3} + 1}{2\sqrt{2} - \sqrt{3}}\\\\\\= \dfrac{\sqrt{3} + 1}{2\sqrt{2} - \sqrt{3}} \times \dfrac{2\sqrt{2} + \sqrt{3}}{2\sqrt{2} + \sqrt{3}}\\\\\\= \dfrac{2\sqrt{6} + 3 + 2\sqrt{2} + \sqrt{3}}{(2\sqrt{2})^2 - (\sqrt{3})^2}\\\\\\= \dfrac{2\sqrt{2} + \sqrt{3} + 2\sqrt{6} + 3}{8 - 3}\\\\\\= \bold{\dfrac{2\sqrt{2} + \sqrt{3} + 2\sqrt{6} + 3}{5}}$