Question

What is 3+root 2 upon root 3 +1

Answer 1

Answer:

I hope it is correct answer

Answer 2

Answer:

$=\frac{3\sqrt{3}-3+\sqrt{6}-\sqrt{2} }{2}$

Step-by-step explanation:

Solving,

$\frac{3+\sqrt{2} }{\sqrt{3}+1 }$

Rationalising,

$\frac{3+\sqrt{2} }{\sqrt{3}+1 } *\frac{\sqrt{3}-1 }{\sqrt{3}-1 }$

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

$=\frac{3\sqrt{3}-3+\sqrt{6}-\sqrt{2} }{(\sqrt{3} )^{2} -1^2}$

$=\frac{3\sqrt{3}-3+\sqrt{6}-\sqrt{2} }{3 -1}$

$=\frac{3\sqrt{3}-3+\sqrt{6}-\sqrt{2} }{2}$