Question

Search root 2 + 1 upon root 2 minus 1 whole square

Answer 1

$3+\sqrt{2}$

Answer:

Step-by-step explanation:

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

Multiply by the $\sqrt{2} + 1$ in denominator & numerator.

Then

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

After simplification

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

Answer