《2025年全国重点高中提前招生同步强化全真试卷八年级数学上册》

11. (北京四中自主招生)多项式$6x^{3} - 11x^{2} + x + 4$可分解为$(x - 1)(3x - 4)(2x + 1)$.

答案： 11.$(x - 1)(3x - 4)(2x + 1)$提示:由系数和为零可知有因式$x - 1$.

12. (2024·重庆)我们规定:若一个正整数$A$能写成$m^{2} - n$,其中$m$与$n$都是两位数,且$m$与$n$的十位数字相同,个位数字之和为$8$,则称$A$为“方减数”,并把$A$分解成$m^{2} - n$的过程,称为“方减分解”. 例如:因为$602 = 25^{2} - 23,25$与$23$的十位数字相同,个位数字$5$与$3$的和为$8$,所以$602$是“方减数”,$602$分解成$602 = 25^{2} - 23$的过程就是“方减分解”. 按照这个规定,最小的“方减数”是82. 把一个“方减数”$A$进行“方减分解”,即$A = m^{2} - n$,将$m$放在$n$的左边组成一个新的四位数$B$,若$B$除以$19$余数为$1$,且$2m + n = k^{2}$($k$为整数),则满足条件的正整数$A$为4564.

答案： 12.824564提示:①设$m = 10a + b$，则$n = 10a + 8 - b(1\leqslant a\leqslant9$，$0\leqslant b\leqslant8)$，由题意得：$m^{2}-n=(10a + b)^{2}-(10a + 8 - b)=100a^{2}+20ab + b^{2}-10a - 8 + b$，$\because1\leqslant a\leqslant9$，$\therefore$要使“方减数”最小，需$a = 1$，$\therefore m = 10 + b$，$n = 18 - b$，$\therefore m^{2}-n=(10 + b)^{2}-(18 - b)=100 + 20b + b^{2}-18 + b=b^{2}+21b + 82$，当$b = 0$时，$m^{2}-n$最小为$82$；②设$m = 10a + b$，则$n = 10a + 8 - b(1\leqslant a\leqslant9$，$0\leqslant b\leqslant8)$，$\therefore B = 1000a + 100b + 10a+8 - b=1010a + 99b + 8$，$\because B$除以$19$余数为$1$，$\therefore 1010a + 99b + 7$能被$19$整除，$\therefore \frac{B - 1}{19}=53a + 5b+\frac{3a + 4b + 7}{19}$为整数，又$2m + n=k^{2}$（$k$为整数），$\therefore 2(10a + b)+10a + 8 - b=30a + b + 8$是完全平方数，$\because1\leqslant a\leqslant9$，$0\leqslant b\leqslant8$，$\therefore 30a + b + 8$最小为$49$，最大为$256$，即$7\leqslant k\leqslant16$，设$3a + 4b + 7 = 19t$，$t$为正整数，则$1\leqslant t\leqslant3$，(I)当$t = 1$时，$3a + 4b = 12$，则$b = 3-\frac{3}{4}a$，$30a + b + 8 = 30a+3-\frac{3}{4}a + 8=\frac{117}{4}a + 11$是完全平方数，又$1\leqslant a\leqslant9$，$0\leqslant b\leqslant8$，此时无整数解，(II)当$t = 2$时，$3a + 4b = 31$，则$b=\frac{31 - 3a}{4}$，$30a + b + 8 = 30a+\frac{31 - 3a}{4}+8=\frac{117}{4}a+\frac{63}{4}$是完全平方数，又$1\leqslant a\leqslant9$，$0\leqslant b\leqslant8$，此时无整数解，(III)当$t = 3$时，$3a + 4b = 50$，则$b=\frac{50 - 3a}{4}$，$30a + b + 8 = 30a+\frac{50 - 3a}{4}+8=\frac{117}{4}a+\frac{82}{4}$是完全平方数，若$a = 6$，$b = 8$，则$3a + 4b + 7 = 57 = 19×3$，$30×6+8 + 8 = 196 = 14^{2}$，$\therefore t = 3$，$k = 14$，此时$m = 10a + 8 = 68$，$n = 10a + 8 - b = 60$，$\therefore A = 68^{2}-60 = 4564$，故答案为$82$，$4564$.

13. (全国初中数学联赛)已知整数$a,b,c$满足不等式$a^{2} + 2b^{2} + c^{2} + 211 ＜ ab + 28b + 20c$,则$a + b - c = $2.

答案： 13.2提示:由已知得$a^{2}+2b^{2}+c^{2}+212 - 1＜ab + 28b + 20c$，$\therefore a^{2}+2b^{2}+c^{2}+212 - ab - 28b - 20c＜1$，$\therefore a^{2}-ab+\frac{1}{4}b^{2}+\frac{7}{4}b^{2}-28b + 112 + c^{2}-20c + 100＜1$，$\therefore (a-\frac{1}{2}b)^{2}+\frac{7}{4}(b - 8)^{2}+(c - 10)^{2}＜1$，$\because (a-\frac{1}{2}b)^{2}+\frac{7}{4}(b - 8)^{2}+(c - 10)^{2}\geqslant0$，$a$，$b$，$c$为整数，$\therefore (a-\frac{1}{2}b)^{2}+\frac{7}{4}(b - 8)^{2}+(c - 10)^{2}=0$，$\therefore\begin{cases}a-\frac{1}{2}b=0\\b - 8 = 0\\c - 10 = 0\end{cases}$，解得$\begin{cases}b = 8\\a = 4\\c = 10\end{cases}$，$\therefore a + b - c = 4 + 8 - 10 = 2$.故答案为$2$.

14. (天津市竞赛)已知四个实数$a,b,c,d$且$a \neq b,c \neq d$,若四个关系式$a^{2} + ac = 2,b^{2} + bc = 2,c^{2} + ac = 4,d^{2} + ad = 4$同时成立,则$6a + 2b + 3c + 2d$的值等于0.

答案： 14.0提示:由已知得$(a^{2}+ac)-(b^{2}+bc)=0$，$(c^{2}+ac)-(d^{2}+ad)=0$，即$(a - b)(a + b + c)=0$，$(c - d)(a + c + d)=0$，$\because a\neq b$，$c\neq d$，$\therefore a + b + c = 0$，$a + c + d = 0$，可得$b = d = -(a + c)$，又$(a^{2}+ac)+(c^{2}+ac)=2 + 4 = 6$，即$a^{2}+2ac + c^{2}=6$，$(a + c)^{2}=6$，$(a^{2}+ac)-(c^{2}+ac)=2 - 4 = -2$，$(a + c)(a - c)= - 2$，$\therefore (a + c)=\pm\sqrt{6}$，$a - c=\pm\frac{\sqrt{6}}{3}$，即$a=\frac{\sqrt{6}}{3}$，$c=\frac{2\sqrt{6}}{3}$或$a=-\frac{\sqrt{6}}{3}$，$c=-\frac{2\sqrt{6}}{3}$，$\therefore 6a + 2b + 3c + 2d=6a + 3c + 4b=6a + 3c - 4(a + c)=2a - c = 0$.

15. 分解因式:(1)(“创新杯”全国数学邀请赛)$x^{3} - ax^{2} - 2ax + a^{2} - 1$.
(2)(上海中学自主招生)$6x^{3} - 11x^{2} + x + 4$.
(3)(衡阳县自主招生)$x^{3} - 3x^{2} - 6x + 8$.

答案： 15.
(1)把$a$看作主元，变形得，原式$=a^{2}-ax^{2}-2ax + x^{2}-1=a^{2}-(x^{2}+2x)a+(x - 1)(x^{2}+x + 1)=[a-(x - 1)][a-(x^{2}+x + 1)]$.
(2)原式$=(6x^{3}-6x^{2})-(5x^{2}-x - 4)=6x^{3}(x - 1)-(5x + 4)(x - 1)=(x - 1)(6x^{2}-5x - 4)=(x - 1)(3x - 4)(2x + 1)$.
(3)原式$=x^{3}(x - 4)+(x - 4)(x - 2)=(x - 4)(x^{2}+x - 2)=(x - 4)(x + 2)(x - 1)$.

16. (南充市顺庆区自主招生改编)设实数$a,b,c$满足$a^{3}(a - 1) + b^{3}(b - 1) + c^{3}(c - 1) = a^{2}(a - 1) + b^{2}(b - 1) + c^{2}(c - 1)$,求满足条件的所有$a,b,c$的值.

答案： 16.解:由已知得$a^{3}(a - 1)-a^{2}(a - 1)+b^{3}(b - 1)-b^{2}(b - 1)+c^{3}(c - 1)-c^{2}(c - 1)=0$.等式左边因式分解得，$a^{2}(a - 1)^{2}+b^{2}(b - 1)^{2}+c^{2}(c - 1)^{2}=0$，又$\because a^{2}(a - 1)^{2}\geqslant0$，$b^{2}(b - 1)^{2}\geqslant0$，$c^{2}(c - 1)^{2}\geqslant0$，$\therefore a^{2}(a - 1)^{2}=0$，$b^{2}(b - 1)^{2}=0$，$c^{2}(c - 1)^{2}=0$，$\therefore a = 0$或$a = 1$；$b = 0$或$b = 1$；$c = 0$或$c = 1$，$\therefore\begin{cases}a = 0\\b = 0\\c = 0\end{cases}$，$\begin{cases}a = 0\\b = 0\\c = 1\end{cases}$，$\begin{cases}a = 0\\b = 1\\c = 0\end{cases}$，$\begin{cases}a = 0\\b = 1\\c = 1\end{cases}$，$\begin{cases}a = 1\\b = 0\\c = 0\end{cases}$，$\begin{cases}a = 1\\b = 0\\c = 1\end{cases}$，$\begin{cases}a = 1\\b = 1\\c = 0\end{cases}$，$\begin{cases}a = 1\\b = 1\\c = 1\end{cases}$.