Для каждого значения а решите уравнение: (x-a)/(x^2-4x+3)=0.

\[\frac{x - a}{x^{2} - 4x + 3} = 0;\ \ \ x \neq 3,\ x \neq 1\]

\[x^{2} - 4x + 3 = 0\]

\[x_{1} + x_{2} = 4;\ \ \ x_{1} \cdot x_{2} = 3\]

\[x_{1} = 3,\ \ x_{2} = 1.\]

\[x^{2} - 4x + 3 = (x - 3)(x - 1)\]

\[x - a = 0\ \]

\[\ x = a.\]

\[1)\ \ a = 3;\ \ \]

\[a = 1 \Longrightarrow нет\ решения.\]

\[2)\ \ a \neq 3;\ \ a \neq 1 \Longrightarrow x = a.\]