Question

ПОМОГИТЕ!!!!
Найдите $\frac{g(4-x)}{g(4+x)}$, если $g(x)=\sqrt[7]{x(8-x)}$ , при IxI≠4

Answer 1

$\displaystyle g(x)=\sqrt[7]{x(8-x)}\\\\\frac{g(4-x)}{g(4+x)}=\frac{\sqrt[7]{(4-x)(8-(4-x))}}{\sqrt[7]{(4+x)(8-(4+x))}}=\sqrt[7]{\frac{(4-x)(8-4+x)}{(4+x)(8-4-x)}}=\\\\\\=\sqrt[7]{\frac{(4-x)(4+x)}{(4+x)(4-x)}}=\sqrt[7]{1}=\boxed{1}$