解:
1. 化简原式:
$\begin{aligned}&(2a^2 - 4a + 1) - (-3a^2 + 2a - 5) =& 2a^2 - 4a + 1 + 3a^2 - 2a + 5 \\=& 5a^2 - 6a + 6\end{aligned}$
变式1:求$2a^2 - 4a + 1$与$-3a^2 + 2a - 5$的差
列式计算:
$\begin{aligned}&(2a^2 - 4a + 1) - (-3a^2 + 2a - 5) =& 2a^2 - 4a + 1 + 3a^2 - 2a + 5 \\=& 5a^2 - 6a + 6\end{aligned}$
变式2:已知$A=2a^2 - 4a + 1,B=-3a^2 + 2a - 5,$求$A-B$
代入计算:
$\begin{aligned}&A - B =& (2a^2 - 4a + 1) - (-3a^2 + 2a - 5) \\=& 2a^2 - 4a + 1 + 3a^2 - 2a + 5 \\=& 5a^2 - 6a + 6\end{aligned}$
变式3:已知$A=2a^2 - 4a + 1,B=-3a^2 + 2a - 5,$其中$a=-1,$求$A-B$的值
先化简得$A-B=5a^2 - 6a + 6,$将$a=-1$代入:
$\begin{aligned}&A-B =& 5×(-1)^2 - 6×(-1) + 6 \\=& 5 + 6 + 6 \\=& 17\end{aligned}$