用行列式表示的方程该怎么求根？

一、题目

难度评级：

二、解析

对于这类用行列式表示函数的题目，我们首先要做的就是先看看能否对这个行列式进行化简，花间之后再进行行列式的运算，运算的结果就是该函数的一般表达式。

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

于是可知，$f ( x )$ $=$ $0$ 的根为：

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

综上可知，本 题 应 选 B

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