Question

Алгебра 8 класс. Дам 25 б​

Answer 1

Ответ:

$x=441\\\\\sqrt{256+32\sqrt{x}+x}=\sqrt{16^2+2\cdot 16\cdot \sqrt{x}+(\sqrt{x})^2}=\sqrt{(16+\sqrt{x})^2}=\\\\=|16+\underbrace{\sqrt{x}}_{\geq 0}|=16+\sqrt{x} =16+\sqrt{441}=16+21=37$