Ответ -20 ,
решение: y=kx;
y=-80,
X=4,
4x=-80,
3 2/5=17/5=17:5=3,4
43/30=43:30=1,4(3)
График f - прямая параллельная оси OX со смещением вдоль оси OY на - 2
![log_ \frac{1}{3}x+log_ \frac{1}{3}(4-x)\ \textgreater \ -1; \left \{ {{x\ \textgreater \ 0} \atop {4-x\ \textgreater \ 0}} \right.=\ \textgreater \ 0\ \textless \ x\ \textless \ 4; \\ log_ \frac{1}{3}(x(4-x))\ \textgreater \ -1; 0\ \textless \ \frac{1}{3}\ \textless \ 1=\ \textgreater \ x(4-x)\ \textless \ ( \frac{1}{3} )^{-1}; \\ -x^2+4x\ \textless \ 3; x^2-4x+3\ \textgreater \ 0; f(x)\ \textgreater \ 0;](https://tex.z-dn.net/?f=log_+%5Cfrac%7B1%7D%7B3%7Dx%2Blog_+%5Cfrac%7B1%7D%7B3%7D%284-x%29%5C+%5Ctextgreater+%5C+-1%3B++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextgreater+%5C+0%7D+%5Catop+%7B4-x%5C+%5Ctextgreater+%5C+0%7D%7D+%5Cright.%3D%5C+%5Ctextgreater+%5C+0%5C+%5Ctextless+%5C+x%5C+%5Ctextless+%5C+4%3B++%5C%5C+log_+%5Cfrac%7B1%7D%7B3%7D%28x%284-x%29%29%5C+%5Ctextgreater+%5C+-1%3B+0%5C+%5Ctextless+%5C++%5Cfrac%7B1%7D%7B3%7D%5C+%5Ctextless+%5C+1%3D%5C+%5Ctextgreater+%5C+x%284-x%29%5C+%5Ctextless+%5C+%28+%5Cfrac%7B1%7D%7B3%7D+%29%5E%7B-1%7D%3B+%5C%5C+-x%5E2%2B4x%5C+%5Ctextless+%5C+3%3B+x%5E2-4x%2B3%5C+%5Ctextgreater+%5C+0%3B+f%28x%29%5C+%5Ctextgreater+%5C+0%3B+++++)
найдём нули
![f(x): x^2-4x+3=0; a+b+c=0 =\ \textgreater \ \left[\begin{array}{ccc}x=1&\\\\x= \frac{c}{a}=3 \end{array}\right; \\ (x-1)(x-3)\ \textgreater \ 0;](https://tex.z-dn.net/?f=f%28x%29%3A+x%5E2-4x%2B3%3D0%3B+a%2Bb%2Bc%3D0+%3D%5C+%5Ctextgreater+%5C+++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%3D1%26%5C%5C%5C%5Cx%3D+%5Cfrac%7Bc%7D%7Ba%7D%3D3+%5Cend%7Barray%7D%5Cright%3B+%5C%5C+%28x-1%29%28x-3%29%5C+%5Ctextgreater+%5C+0%3B)
x∈(-∞;1)∨(3;+∞), но по ОДЗ x∈(0;4), поэтому накладывая условие ОДЗ, получаем x∈(0;1)∨(3;4)