Помогите упростить 6 б, пожалуйста.

Решите задачу:

$\sqrt{29+4\sqrt7}+ \frac{27}{1+2\sqrt7} =\sqrt{1+28+2\sqrt{28}}+ \frac{27}{1+2\sqrt7} =\\\\=\sqrt{(1+\sqrt{28})^2}+\frac{27}{1+2\sqrt7}=|1+\sqrt{28}|+\frac{27}{1+2\sqrt7}=\\\\=1+2\sqrt7+\frac{27}{1+2\sqrt7}= \frac{(1+2\sqrt7)^2+27}{1+2\sqrt7} = \frac{1+4\sqrt7+28+27}{1+2\sqrt7} =\\\\= \frac{56+4\sqrt7}{1+2\sqrt7} = \frac{4(14+\sqrt7)}{1+2\sqrt7}$