Alg2.1.11

Simplify $\sqrt{6-\sqrt{7}+\sqrt{27+4\sqrt{35}}}\,$

Readjusting the values,to find out the suare root inside the root

$\sqrt{6-\sqrt{7}+\sqrt{27+2(2)\sqrt{7\cdot5}}}\,$

$\sqrt{6-\sqrt{7}+\sqrt{20+7+2\sqrt{7}(2\sqrt{5})}}\,$

Readjusting the values

$\sqrt{6-\sqrt{7}+\sqrt{(2\sqrt{5})^2+(\sqrt{7})^2+2\sqrt{7}(2\sqrt{5})}}\,$

$\sqrt{6-\sqrt{7}+\sqrt{(2\sqrt{5}+\sqrt{7})^2}}\,$

Simplifying further

$\sqrt{6-\sqrt{7}+2\sqrt{5}+\sqrt{7}}\,$

Simplifying $\sqrt{6+2\sqrt{5}}\,$

Here again applying the same rule like in the earlier steps

$\sqrt{5+1+2\sqrt{5(1)}}\,$

Expressing the values in squares

$\sqrt{(\sqrt{5})^2+1^2+2\sqrt{5(1)}}\,$

This can be written as

$\sqrt{(\sqrt{5}+1)^2}\,$

Hence the result is

$\sqrt{5}+1\,$