Question

(x-x/1-x):3x^4/x^2-2x+1 при x=1/6

Answer 1

$mathtt{(x-frac{x}{1-x}):frac{3x^4}{x^2-2x+1}=frac{x(x-1)+x}{x-1}*frac{x^2-2x+1}{3x^4}=frac{x^2-x+x}{x-1}*frac{(x-1)^2}{3x^4}=frac{x-1}{3x^2};}$

при $mathtt{x=frac{1}{6}}$: $mathtt{frac{frac{1}{6}-1}{3*(frac{1}{6})^2}=frac{frac{1-6}{6}}{3*frac{1}{36}}=frac{-frac{5}{6}}{frac{1}{12}}=-frac{5}{6}:frac{1}{12}=-frac{5}{6}*12=-5*2=-10}$