1. (2025·苏州期末)先化简,再求值:$(a + 2b)·(2b - a)-(a + 2b)^{2}-4ab$,其中$a = 1$,$b=\frac{1}{2025}$.
答案：1. 原式$=4b^{2}-a^{2}-(a^{2}+4ab+4b^{2})-4ab=4b^{2}-a^{2}-a^{2}-4ab-4b^{2}-4ab=-2a^{2}-8ab$，当$a=1$，$b=\frac{1}{2025}$时，原式$=-2×1^{2}-8×1×\frac{1}{2025}=-2-\frac{8}{2025}=-2\frac{8}{2025}$。
2. (2025·无锡期末)先化简,再求值:$(2x - 1)^{2}-(3x + 1)(3x - 1)+4x(x - 1)$,其中$x^{2}+8x - 3 = 0$.
答案：2. 原式$=4x^{2}-4x+1-(9x^{2}-1)+4x^{2}-4x=4x^{2}-4x+1-9x^{2}+1+4x^{2}-4x=-x^{2}-8x+2$，$\because x^{2}+8x-3=0$，$\therefore -x^{2}-8x=-3$，$\therefore$原式$=-3+2=-1$。
3. (2025·淮安期末)先化简,再求值:$(-\frac{3}{2}x+\frac{1}{3}y^{2})(2x-\frac{2}{3}y)-\frac{1}{3}y^{2}(x-\frac{2}{3}y)+3x^{2}$,其中$x$,$y$满足$|xy - 2|+(y + 2)^{2}=0$.
答案：3. 原式$=-3x^{2}+xy+\frac{2}{3}xy^{2}-\frac{2}{9}y^{3}-\frac{1}{3}xy^{2}+\frac{2}{9}y^{3}+3x^{2}=xy+\frac{xy^{2}}{3}$。$\because |xy - 2|+(y + 2)^{2}=0$，$\therefore xy - 2 = 0$且$y + 2 = 0$，$\therefore xy = 2$，$y = - 2$，$\therefore xy^{2}=-4$，原式$=2-\frac{4}{3}=\frac{2}{3}$。
解析：
原式$=(-\frac{3}{2}x)(2x)+(-\frac{3}{2}x)(-\frac{2}{3}y)+\frac{1}{3}y^{2}(2x)+\frac{1}{3}y^{2}(-\frac{2}{3}y)-\frac{1}{3}y^{2}· x+\frac{1}{3}y^{2}·\frac{2}{3}y+3x^{2}$
$=-3x^{2}+xy+\frac{2}{3}xy^{2}-\frac{2}{9}y^{3}-\frac{1}{3}xy^{2}+\frac{2}{9}y^{3}+3x^{2}$
$=(-3x^{2}+3x^{2})+xy+(\frac{2}{3}xy^{2}-\frac{1}{3}xy^{2})+(-\frac{2}{9}y^{3}+\frac{2}{9}y^{3})$
$=xy+\frac{1}{3}xy^{2}$
$\because |xy - 2|+(y + 2)^{2}=0$，
$\therefore xy - 2 = 0$且$y + 2 = 0$，
$\therefore xy = 2$，$y = - 2$，
$\therefore x=\frac{xy}{y}=\frac{2}{-2}=-1$，
$\therefore xy^{2}=(-1)×(-2)^{2}=-4$，
原式$=2+\frac{1}{3}×(-4)=2-\frac{4}{3}=\frac{2}{3}$
$=-3x^{2}+xy+\frac{2}{3}xy^{2}-\frac{2}{9}y^{3}-\frac{1}{3}xy^{2}+\frac{2}{9}y^{3}+3x^{2}$
$=(-3x^{2}+3x^{2})+xy+(\frac{2}{3}xy^{2}-\frac{1}{3}xy^{2})+(-\frac{2}{9}y^{3}+\frac{2}{9}y^{3})$
$=xy+\frac{1}{3}xy^{2}$
$\because |xy - 2|+(y + 2)^{2}=0$，
$\therefore xy - 2 = 0$且$y + 2 = 0$，
$\therefore xy = 2$，$y = - 2$，
$\therefore x=\frac{xy}{y}=\frac{2}{-2}=-1$，
$\therefore xy^{2}=(-1)×(-2)^{2}=-4$，
原式$=2+\frac{1}{3}×(-4)=2-\frac{4}{3}=\frac{2}{3}$
4. 已知代数式$(ax - 3)(2x + 4)-x^{2}-b$化简后,不含有$x^{2}$项和常数项.
(1)求$a$,$b$的值;
(2)求$(b - a)(-a - b)+(-a - b)^{2}-a(2a + b)$的值.
(1)求$a$,$b$的值;
(2)求$(b - a)(-a - b)+(-a - b)^{2}-a(2a + b)$的值.
答案：4. (1)$(ax - 3)(2x + 4)-x^{2}-b=2ax^{2}+4ax - 6x - 12 - x^{2}-b=(2a - 1)x^{2}+(4a - 6)x+( - 12 - b)$。$\because$代数式$(ax - 3)(2x + 4)-x^{2}-b$化简后，不含有$x^{2}$项和常数项，$\therefore 2a - 1 = 0$，$-12 - b = 0$，$\therefore a=\frac{1}{2}$，$b=-12$。
(2)$\because a=\frac{1}{2}$，$b=-12$，$\therefore (b - a)(-a - b)+(-a - b)^{2}-a(2a + b)=a^{2}-b^{2}+a^{2}+2ab + b^{2}-2a^{2}-ab=ab=\frac{1}{2}×(-12)=-6$。
(2)$\because a=\frac{1}{2}$，$b=-12$，$\therefore (b - a)(-a - b)+(-a - b)^{2}-a(2a + b)=a^{2}-b^{2}+a^{2}+2ab + b^{2}-2a^{2}-ab=ab=\frac{1}{2}×(-12)=-6$。
5. 对于任意有理数$a$,$b$,$c$,$d$,我们规定符号$(a,b)\otimes(c,d)=ad - bc$,例如:$(1,3)\otimes(2,4)=1×4 - 3×2=-2$.
(1)$(-2,3)\otimes(4,5)$的值为
(2)求$(3a + 1,a - 2)\otimes(a + 2,a - 3)$的值,其中$a^{2}-4a + 1 = 0$.
(1)$(-2,3)\otimes(4,5)$的值为
-22
;(2)求$(3a + 1,a - 2)\otimes(a + 2,a - 3)$的值,其中$a^{2}-4a + 1 = 0$.
答案：5. (1)$-22$ 解析：$(-2,3)\otimes(4,5)=-2×5 - 3×4=-10 - 12=-22$。
(2)$(3a + 1,a - 2)\otimes(a + 2,a - 3)=(3a + 1)(a - 3)-(a - 2)·(a + 2)=3a^{2}-9a + a - 3-(a^{2}-4)=3a^{2}-9a + a - 3 - a^{2}+4=2a^{2}-8a + 1$。$\because a^{2}-4a + 1 = 0$，$\therefore a^{2}=4a - 1$。$\therefore (3a + 1,a - 2)\otimes(a + 2,a - 3)=2(4a - 1)-8a + 1=-1$。
(2)$(3a + 1,a - 2)\otimes(a + 2,a - 3)=(3a + 1)(a - 3)-(a - 2)·(a + 2)=3a^{2}-9a + a - 3-(a^{2}-4)=3a^{2}-9a + a - 3 - a^{2}+4=2a^{2}-8a + 1$。$\because a^{2}-4a + 1 = 0$，$\therefore a^{2}=4a - 1$。$\therefore (3a + 1,a - 2)\otimes(a + 2,a - 3)=2(4a - 1)-8a + 1=-1$。