# 用行列式表示的方程该怎么求根？

## 二、解析

\begin{aligned} f ( x ) \\ \\ & = \left| \begin{array} { c c c c } x – 2 & x – 1 & x – 2 & x – 3 \\ 2 x – 2 & 2 x – 1 & 2 x – 2 & 2 x – 3 \\ 3 x – 3 & 3 x – 2 & 4 x – 5 & 3 x – 5 \\ 4 x & 4 x – 3 & 5 x – 7 & 4 x – 3 \end{array} \right| \\ \\ & = \left| \begin{array} { c c c c } x – 2 & 1 & 0 & – 1 \\ 2 x – 2 & 1 & 0 & – 1 \\ 3 x – 3 & 1 & x – 2 & – 2 \\ 4 x & – 3 & x – 7 & – 3 \end{array} \right| \\ \\ & = \left| \begin{array} { c c c c } \textcolor{springgreen}{x – 2} & \textcolor{springgreen}{1} & 0 & 0 \\ \textcolor{springgreen}{2 x – 2} & \textcolor{springgreen}{1} & 0 & 0 \\ 3 x – 3 & 1 & \textcolor{orangered}{x – 2} & \textcolor{orangered}{- 1} \\ 4 x & – 3 & \textcolor{orangered}{x – 7} & \textcolor{orangered}{- 6} \end{array} \right| \\ \\ & = \left| \begin{array} { l l } \textcolor{springgreen}{x – 2} & \textcolor{springgreen}{1} \\ \textcolor{springgreen}{2 x – 2} & \textcolor{springgreen}{1} \end{array} \right| \cdot \left| \begin{array} { l l } \textcolor{orangered}{x – 2} & \textcolor{orangered}{- 1} \\ \textcolor{orangered}{x – 7} & \textcolor{orangered}{- 6} \end{array} \right| \\ \\ & = (x-2-2x+2)(-6x+12+x-7) \\ \\ & = – x \cdot ( 5 – 5 x ) \end{aligned}

$$\begin{cases} x = 0 \\ x = 1 \end{cases}$$