《2025年课时训练七年级数学下册苏科版江苏人民出版社》

22. (4分)对于任意有理数$a$,$b$,$c$,$d$,我们规定符号$(a,b)(c,d) = ad - bc$,例如:$(1,3)(2,4) = 1 \times 4 - 2 \times 3 = -2$.
(1)求$(-2,3)(4,5)$的值为 -22;
(2)求$(3a + 1,a - 2)(a + 2,a - 3)$的值,其中$a^{2} - 4a + 1 = 0$.
解:(2)$(3a + 1,a - 2)(a + 2,a - 3) = (3a + 1)(a - 3) - (a - 2)(a + 2) = 3a^{2} - 9a + a - 3 - (a^{2} - 4) = 3a^{2} - 9a + a - 3 - a^{2} + 4 = 2a^{2} - 8a + 1$.
$\because a^{2} - 4a + 1 = 0$,
$\therefore a^{2} = 4a - 1$,
$\therefore(3a + 1,a - 2)(a + 2,a - 3) = 2(4a - 1) - 8a + 1 = -1$.

答案：
(1)$-22$
(2)$(3a + 1,a - 2)(a + 2,a - 3)=(3a + 1)(a - 3)-(a - 2)(a + 2)=3a^{2}-9a + a - 3-(a^{2}-4)=3a^{2}-9a + a - 3 - a^{2}+4 = 2a^{2}-8a + 1$.
$\because a^{2}-4a + 1 = 0$，
$\therefore a^{2}=4a - 1$，
$\therefore(3a + 1,a - 2)(a + 2,a - 3)=2(4a - 1)-8a + 1=-1$.

23. (6分)小红准备完成题目:计算$(x^{2} ★ x + 2)(x^{2} - x)$,她发现第一个因式的一次项系数被墨水遮挡住了.
(1)她把被遮住的一次项系数猜成3,请你完成计算:$(x^{2} + 3x + 2)(x^{2} - x)$.
(2)老师说:"你猜错了,这个题目的正确答案是不含三次项的."请通过计算说明原题中被遮住的一次项系数是多少?
解:(1)$(x^{2} + 3x + 2)(x^{2} - x) = x^{4} - x^{3} + 3x^{3} - 3x^{2} + 2x^{2} - 2x = x^{4} + 2x^{3} - x^{2} - 2x$.
(2)设一次项系数为$a$,
$(x^{2} + ax + 2)(x^{2} - x) = x^{4} - x^{3} + ax^{3} - ax^{2} + 2x^{2} - 2x = x^{4} + (a - 1)x^{3} + (2 - a)x^{2} - 2x$.
$\because$答案是不含三次项的,
$\therefore a - 1 = 0$,$\therefore a = 1$.

答案：
(1)$(x^{2}+3x + 2)(x^{2}-x)=x^{4}-x^{3}+3x^{3}-3x^{2}+2x^{2}-2x = x^{4}+2x^{3}-x^{2}-2x$.
(2)设一次项系数为 $a$，
$(x^{2}+ax + 2)(x^{2}-x)=x^{4}-x^{3}+ax^{3}-ax^{2}+2x^{2}-2x = x^{4}+(a - 1)x^{3}+(2 - a)x^{2}-2x$.
$\because$答案是不含三次项的，
$\therefore a - 1 = 0$，$\therefore a = 1$.