a)              \(x^2-5x+4=0\)

\(\Leftrightarrow\)\(x^2-x-4x+4=0\)

\(\Leftrightarrow\)\(x\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-1=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=4\end{cases}}\)

Vậy tổng các giá trị nguyên của x thỏa mãn là:

                \(1+4=5\)

Ta có : \(\frac{x+4}{5}=\frac{20}{x+4}\)

\(\Rightarrow\left(x+4\right).\left(x+4\right)=5.20\)

\(\Rightarrow\left(x+4\right).\left(x+4\right)=100\)

\(\Rightarrow\left(x+4\right)^2=\left(\pm10\right)^2\)

\(\Leftrightarrow\left(x+4\right)=10\)

\(\Rightarrow x=10-4\)

\(\Rightarrow x=6\)

hoặc 

\(\Leftrightarrow\left(x+4\right)=-10\)

\(\Rightarrow x=-10-4\)

\(\Rightarrow x=-14\)

\(\frac{x+4}{20}=\frac{5}{x+4}\)

\(\left(x+4\right)^2=100\)

\(\left(x+4\right)^2=\left(\pm10\right)^2\)

\(x+4=\pm10\)

                \(x=10-4\)

                \(x=6\)

                \(x=-10-4\)
                \(x=-14\)

Vậy \(x\in\left\{-14;6\right\}\)

\(\frac{x+4}{20}\) = \(\frac{5}{x+4}\)

\(\Rightarrow\) (x+4)(x+4) = 5.20 =100

    (x+4)² =10²

             x=6

          x= -14

Vậy :............