Question

$\sqrt{(4-\sqrt{29})^{2}$

Answer 1

Ответ:

$\sqrt{\left(4-\sqrt{29}\right)^2}\\=\sqrt{45-8\sqrt{29}}\\=\sqrt{29-8\sqrt{29}+16}\\=\sqrt{1\cdot \:29-8\sqrt{29}+16}\\\\=\sqrt{\left(\sqrt{1}\right)^2\left(\sqrt{29}\right)^2-8\sqrt{29}+\left(\sqrt{16}\right)^2}=\sqrt{1^2\left(\sqrt{29}\right)^2-2\cdot \:1\cdot \:4\sqrt{29}+4^2}\\=\sqrt{\left(1\cdot \sqrt{29}-4\right)^2}\\=\sqrt{29}-4$

Пошаговое объяснение: