Хелп хелп 30 баллов

$\displaystyle\tt \frac{3x(2-x)^{-1}}{3-\bigg(\cfrac{2-x}{3x}\bigg)^{-1}}=\frac{3x(2-x)^{-1}}{3-\cfrac{3x}{2-x}}=\frac{3x}{\cfrac{3(2-x)-3x}{2-x}\cdot(2-x)}=\\\\\\=\frac{3x}{6-3x-3x}=\frac{3x}{6-6x}=\frac{3x}{3(2-2x)}=\frac{x}{2-2x}$

при х = 4/7:

$\displaystyle\tt\frac{x}{2-2x}=\frac{\cfrac{4}{7} }{2-2\cdot\cfrac{4}{7} }=\frac{\cfrac{4}{7} }{2-\cfrac{8}{7} }= \frac{\cfrac{4}{7} }{\cfrac{6}{7} }=\frac{4}{6} =\frac{2}{3}$