《2025年新课程学习指导南方出版社八年级数学上册华师大版》

1. (基础题·运算能力)计算:
(1)$(2x - 1)(x - 3) - 2(x - 1)^2$
(2)$(2x - 3y)^2 - (2x + 3y)^2$
(3)$2025^2 - 2024^2$ (4)$999^2 + 2×99 + 1^2$

答案：
(1)
$\begin{aligned}&(2x - 1)(x - 3) - 2(x - 1)^2\\=&2x^2 - 6x - x + 3 - 2(x^2 - 2x + 1)\\=&2x^2 - 7x + 3 - 2x^2 + 4x - 2\\=&-3x + 1\end{aligned}$
(2)
$\begin{aligned}&(2x - 3y)^2 - (2x + 3y)^2\\=&(4x^2 - 12xy + 9y^2) - (4x^2 + 12xy + 9y^2)\\=&4x^2 - 12xy + 9y^2 - 4x^2 - 12xy - 9y^2\\=&-24xy\end{aligned}$
(3)
$\begin{aligned}&2025^2 - 2024^2\\=&(2025 - 2024)(2025 + 2024)\\=&1×4049\\=&4049\end{aligned}$
(4)
$\begin{aligned}&999^2 + 2×999×1 + 1^2\\=&(999 + 1)^2\\=&1000^2\\=&1000000\end{aligned}$

2. (综合题·因式分解)分解因式:
(1)$x(x - 6) + 9$ (2)$x^4 - 1$
(3)$3ax^2 - 3ay^2$ (4)$25x^2 - 36$
(5)$x^2(a - b) + 9(b - a)$ (6)$x^2 + 4(x + 2) - 4$

答案：
(1)
$\begin{aligned}x(x - 6) + 9 \\= x^{2}-6x + 9\\=(x - 3)^{2}\end{aligned}$
(2)
$\begin{aligned}x^{4}-1\\=(x^{2}+1)(x^{2}-1)\\=(x^{2}+1)(x + 1)(x - 1)\end{aligned}$
(3)
$\begin{aligned}3ax^{2}-3ay^{2}\\=3a(x^{2}-y^{2})\\=3a(x + y)(x - y)\end{aligned}$
(4)
$\begin{aligned}25x^{2}-36\\=(5x)^{2}-6^{2}\\=(5x + 6)(5x - 6)\end{aligned}$
(5)
$\begin{aligned}x^{2}(a - b)+9(b - a)\\=x^{2}(a - b)-9(a - b)\\=(a - b)(x^{2}-9)\\=(a - b)(x + 3)(x - 3)\end{aligned}$
(6)
$\begin{aligned}x^{2}+4(x + 2)-4\\=x^{2}+4x+8 - 4\\=x^{2}+4x + 4\\=(x + 2)^{2}\end{aligned}$

3. (经典题·数学计算)化简求值:
$(2x + 3y)(2x - 3y) - (x - 2y)^2 - (6x^2y + 2y^3) ÷ (-\frac{1}{2}y)$,其中$x = - 3,y = \frac{1}{3}$.

答案： 130

4. (经典题·数学计算)已知$a^2 + b^2 + 2a - 4b + 5 = 0$,求$2a^2 + 4b - 2$的值.

答案： 已知$a^2 + b^2 + 2a - 4b + 5 = 0$，将等式左边配方：
$\begin{aligned}a^2 + 2a + b^2 - 4b + 5&=(a^2 + 2a + 1) + (b^2 - 4b + 4)\\&=(a + 1)^2 + (b - 2)^2\end{aligned}$
则$(a + 1)^2 + (b - 2)^2 = 0$。
因为平方数非负，所以$a + 1 = 0$且$b - 2 = 0$，解得$a = -1$，$b = 2$。
将$a = -1$，$b = 2$代入$2a^2 + 4b - 2$：
$\begin{aligned}2a^2 + 4b - 2&=2×(-1)^2 + 4×2 - 2\\&=2×1 + 8 - 2\\&=2 + 8 - 2\\&=8\end{aligned}$

5. (经典题·数学计算)先化简,再求值:$(2x + y)^2 - (2x + y)(2x - y) - 2y(x + y)$,其中$x = (\frac{1}{2})^{2025},y = 2^{2024}$.

答案： 1