2006-II-06

答案：B
I 並不是恒等式。設 $x=1$，

$\begin{array}{rcl}
\text{左方} & = & (1)^2 -4 \\
& = & -3 \\
\text{右方} & = & 0 \\
\therefore \text{左方} & \neq & \text{右方}
\end{array}$

II 並不是恒等式。設 $x=1$，

$\begin{array}{rcl}
\text{左方} & = & (1)^2 -4 \\
& = & -3 \\
\text{右方} & = & ((1)-2)^2 \\
& = & 1 \\
\therefore \text{左方} & \neq & \text{右方}
\end{array}$

III 是一恒等式。對於任何數值 $x$，

$\begin{array}{rcl}
\text{右方} & = & (x+2)(x-2) \\
& = & x^2 +2x -2x -4 \\
& = & x^2 -4 \\
\text{左方} & = & x^2 – 4 \\
\therefore \text{左方} & = & \text{右方}
\end{array}$