《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》

1. (2022·深圳市第九届“鹏程杯”邀请赛)$a^{4}(b^{2}-c^{2})+b^{4}(c^{2}-a^{2})+c^{4}(a^{2}-b^{2})$的一个因式为 (D)

A.$(a + b)^{2}$
B.$(a - b)^{2}$
C.$a^{2} + b^{2}$
D.$a^{2} - b^{2}$
E.以上都不对

答案： 1.D提示:原式$=a^{4}b^{2}-a^{4}c^{2}+b^{2}c^{2}-a^{2}b^{4}+a^{4}c^{2}-b^{2}c^{4}=(a^{4}b^{2}-a^{2}b^{4}-a^{2}c^{2}+a^{4}c^{2})+(b^{2}c^{2}-b^{2}c^{4})=a^{2}(a^{2}b^{2}-a^{2}c^{2}-b^{2}+c^{2})+b^{2}c^{2}(b^{2}-c^{2})=a^{2}[a^{2}(b^{2}-c^{2})-(b^{2}-c^{2})]+b^{2}c^{2}(b^{2}-c^{2})=a^{2}(b^{2}-c^{2})(a^{2}-1)+b^{2}c^{2}(b^{2}-c^{2})=(b^{2}-c^{2})(a^{2}-a^{2}+b^{2}c^{2})=(b^{2}-c^{2})[a^{2}(b^{2}-c^{2})-(b^{2}+c^{2})b^{2}+b^{2}c^{2}]=(b^{2}-c^{2})[a^{2}(b^{2}-c^{2})-b^{4}-b^{2}c^{2}+b^{2}c^{2}]=(b^{2}-c^{2})[a^{2}(b^{2}-c^{2})-a^{2}c^{2}+b^{2}c^{2}-a^{2}c^{2}]=(b^{2}-c^{2})[(a^{2}-a^{2}c^{2})+(b^{2}c^{2}-a^{2}b^{2})]=(b^{2}-c^{2})[a^{2}(a^{2}-c^{2})-b^{2}(a^{2}-c^{2})]=(b^{2}-c^{2})(a^{2}-b^{2})(a^{2}-c^{2})$，$\therefore$原式的一个因式为$a^{2}-b^{2}$.故选D.

2. (江苏自主招生)已知$a,b,c$是正整数,且$a ＞ b,a^{2} - ab - ac + bc = 11$,则$a - c$等于 (D)

A.$-1$
B.$-1$或$-11$
C.$1$
D.$1$或$11$

答案： 2.D提示:原式$=(a^{2}-ab)-(ac-bc)=a(a-b)-c(a-b)=(a-b)(a-c)=11$.因为$a＞b$，$a$，$b$，$c$是正整数，所以$a - b＞0$，$a - b = 11$或$1$，$a - c = 1$或$11$，故$b = 1$，$c = 3$，$d = 4$，$e = 5$，$f = 6$，选D.

3. (2023 秋·重庆九龙坡区期末)已知六元方程$a + b + c + d + e + f = b^{2} - a^{2} + d^{2} - c^{2} + f^{2} - e^{2}$,满足$a ＜ b ＜ c ＜ d ＜ e ＜ f$,且$a,b,c,d,e,f$为正整数,则下列关于这个六元方程的正整数解的说法中正确的个数为 (D)
①$a = 1,b = 2,c = 3,d = 4,e = 5,f = 6$是该六元方程的一组解;
②连续的六个正整数一定是该六元方程的解;
③若$a ＜ b ＜ c ＜ d ＜ e ＜ f ＜ 10$,则该六元方程有 20 组解;
④若$a + b + c + d + e + f = 23$,则该六元方程有 1 组解.

A.$1$
B.$2$
C.$3$
D.$4$

答案： 3.D提示:$\because a = 1$，$b = 2$，$c = 3$，$d = 4$，$e = 5$，$f = 6$，$\therefore a + b + c + d + e + f = 21$，$b^{2}-a^{2}+d^{2}-c^{2}+f^{2}-e^{2}=21$，$\therefore a + b + c + d + e + f = b^{2}-a^{2}+d^{2}-c^{2}+f^{2}-e^{2}$，$\therefore a = 1$，$b = 2$，$c = 3$，$d = 4$，$e = 5$，$f = 6$是该六元方程的一组解，$\therefore$①正确；设最小的正整数为$n$，那么其余的数为$n + 1$，$n + 2$，$n + 3$，$n + 4$，$n + 5$，$\therefore a + b + c + d + e + f = 6n + 15$，$b^{2}-a^{2}+d^{2}-c^{2}+f^{2}-e^{2}=(n + 1)^{2}-n^{2}+(n + 3)^{2}-(n + 2)^{2}+(n + 5)^{2}-(n + 4)^{2}=6n + 15$，$\therefore a + b + c + d + e + f = b^{2}-a^{2}+d^{2}-c^{2}+f^{2}-e^{2}$，$\therefore$连续的六个正整数一定是该六元方程的解，$\therefore$②正确；$\because a＜b＜c＜d＜e＜f＜10$，连续的正整数解为$1$，$2$，$3$，$4$，$5$，$6$；$2$，$3$，$4$，$5$，$6$，$7$；$3$，$4$，$5$，$6$，$7$，$8$；$4$，$5$，$6$，$7$，$8$，$9$共$4$组；$\because a + b + c + d + e + f = b^{2}-a^{2}+d^{2}-c^{2}+f^{2}-e^{2}$，$\therefore (a + b)+(c + d)+(e + f)=(b - a)(b + a)+(d - c)(d + c)+(f - e)(f + e)$，$\therefore b - a = 1$，$d - c = 1$，$f - e = 1$，$\therefore$不连续正整数解为$1$，$2$，$3$，$4$，$6$，$7$；$1$，$2$，$3$，$4$，$7$，$8$；$1$，$2$，$3$，$4$，$8$，$9$；$1$，$2$，$4$，$5$，$6$，$7$；$1$，$2$，$4$，$5$，$7$，$8$；$1$，$2$，$4$，$5$，$8$，$9$；$2$，$3$，$4$，$5$，$6$，$7$；$2$，$3$，$4$，$5$，$7$，$8$；$2$，$3$，$4$，$5$，$8$，$9$；$2$，$3$，$5$，$6$，$7$，$8$；$2$，$3$，$5$，$6$，$8$，$9$；$3$，$4$，$5$，$6$，$7$，$8$；$2$，$4$，$5$，$6$，$7$，$8$；$1$，$2$，$5$，$6$，$7$，$8$；$1$，$2$，$5$，$6$，$8$，$9$；$1$，$2$，$6$，$7$，$8$，$9$；$2$，$3$，$4$，$5$，$6$，$9$共$16$组，$\therefore$则该六元方程有$20$组解，$\therefore$③正确；$\because a + b + c + d + e + f = 23$，只有$1 + 2 + 3 + 4 + 6 + 7 = 23$，$\therefore$则该六元方程有$1$组解：$a = 1$，$b = 2$，$c = 3$，$d = 4$，$e = 6$，$f = 7$，故④正确.故选D.

4. (2023·宁波自主招生)若实数$x,y$满足$x^{3} - y^{3} + 3xy + 1 = 0$,则$x - y$可能的值 (B)

A.只有$1$个
B.有$2$个
C.多于$2$个但有限
D.有无数个

答案： 4.B提示:$\because x^{3}-y^{3}+3xy + 1 = 0$，$\therefore x^{3}-y^{3}+3xy = -1$①，$\because (x - y)^{3}=x^{3}-y^{3}-3x^{2}y + 3xy^{2}$②，②$-$①得：$-3x^{2}y - 3xy + 3xy^{2}=(x - y)^{3}+1$，$\frac{1}{2}(x - y + 1)[(x + y)^{2}+(x - 1)^{2}+(y + 1)^{2}]=0$，$\therefore x - y + 1 = 0$或$(x - y)^{2}+(x - 1)^{2}+(y + 1)^{2}=0$，当$x - y + 1 = 0$时$x - y = -1$；当$(x - y)^{2}+(x - 1)^{2}+(y + 1)^{2}=0$时，$\because (x + y)^{2}\geqslant0$，$(x - 1)^{2}\geqslant0$，$(y + 1)^{2}\geqslant0$，$\therefore x - 1 = 0$，$y + 1 = 0$，$\therefore x = 1$，$y = -1$，$\therefore x - y = 1 - (-1)=1 + 1 = 2$，综上可知：$x - y$的值有$2$个，为$-1$或$2$，故选B.

5. (北京市竞赛)$a^{4} + 4$分解因式的结果是 (D)

A.$(a^{2} + 2a - 2)(a^{2} - 2a + 2)$
B.$(a^{2} + 2a - 2)(a^{2} - 2a - 2)$
C.$(a^{2} + 2a + 2)(a^{2} - 2a - 2)$
D.$(a^{2} + 2a + 2)(a^{2} - 2a + 2)$

答案： 5.D提示:$a^{2}+4=(a^{2}+4a^{2}+4)-4a^{2}=(a^{2}+2)^{2}-(2a)^{2}=(a^{2}+2a + 2)(a^{2}-2a + 2)$.

6. (江苏省“时代杯”数学应用与创新邀请赛)设正整数$a,b,c ＞ 100$,满足$c^{2} - 1 = a^{2}(b^{2} - 1)$,则$\frac{a}{b}$的最小值是 (C)

A.$\frac{1}{3}$
B.$\frac{1}{2}$
C.$2$
D.$3$

答案： 6.C提示:由已知得$c^{2}=a^{2}(b^{2}-1)+1=a^{2}b^{2}-a^{2}+1＜a^{2}b^{2}$，$\therefore c＜ab$，$\therefore c\leqslant ab - 1$，$\therefore a^{2}b^{2}-a^{2}+1=c^{2}\leqslant(ab - 1)^{2}$，化简，得$a^{2}\geqslant2ab$，$\therefore \frac{a}{b}\geqslant2$，故选C.

7. (“希望杯”竞赛)若$m = 2006^{2} + 2006^{2}×2007^{2} + 2007^{2}$,则$m$ (A)

A.是完全平方数,还是奇数
B.是完全平方数,还是偶数
C.不是完全平方数,但是是奇数
D.不是完全平方数,但是是偶数

答案： 7.A提示:设$n = 2006$，则$n + 1 = 2007$，代入原式，由已知得$m=n^{2}+n^{2}×(n + 1)^{2}+(n + 1)^{2}=n^{2}(n + 1)^{2}+2n(n + 1)+1=[n(n + 1)+1]^{2}$，$\because n(n + 1)$是$2$的倍数，$\therefore n(n + 1)+1$是奇数，$\therefore m$是完全平方数，而且还是奇数，故选A.

8. (湖北省黄冈竞赛)已知$a,b,c$为一个三角形的三边长,则$4b^{2}c^{2} - (b^{2} + c^{2} - a^{2})^{2}$的值为 (A)

A.恒为正
B.恒为负
C.可正可负
D.非负

答案： 8.A提示:原式$=(2bc + b^{2}+c^{2}-a^{2})(2bc - b^{2}-c^{2}+a^{2})=[(b + c)^{2}-a^{2}][a^{2}-(b - c)^{2}]=(b + c + a)(b + c - a)(a + b - c)(a - b + c)$，$\because a$，$b$，$c$是三角形的三边长，$\therefore b + c + a＞0$，$\therefore b + c＞a$，$a + b＞c$，$a + c＞b$，$\therefore 4b^{2}c^{2}-(b^{2}+c^{2}-a^{2})^{2}＞0$.

9. (“学而思杯”中学生理科能力大赛)在$\triangle ABC$中,已知$AB = 37,AC = 58$,在$BC$上有一点$D$使得$AB = AD$,且$D$在$B,C$之间. 若$BD$与$DC$的长度都是整数,则$BD$的长度是22.

答案：
9.22提示:过点$A$作$AH\perp BC$于$H$，则$AC^{2}=AH^{2}+CH^{2}$，$AB^{2}=AH^{2}+BH^{2}$，故$AC^{2}-AB^{2}=CH^{2}-BH^{2}=(CH + BH)(CH - BH)=BC× CD$，$\because AB = 37$，$AC = 58$，$\therefore BC× CD=58^{2}-37^{2}=3×5×7×19$，$\because AC - AB＜BC＜AC + AB$，$\therefore 21＜BC＜95$，$\because BC$为整数，$\therefore BC = 35$或$BC = 57$，若$BC = 35$，则$CD = 3×19 = 57＞BC$，$D$不在$B$，$C$之间，故应舍去，$\therefore$应取$BC = 57$，这时$CD = 35$，$BD = 22$.故答案为$22$.

10. (温州初中数学竞赛)方程$xyz + xy + xz + yz + x + y + z = 2012$的非负整数解有27组.

答案： 10.27提示:原式$=(xyz + yz)+(xy + y)+(xz + z)+(x + 1)-1=yz(x + 1)+y(x + 1)+z(x + 1)+(x + 1)-1=(x + 1)(yz + y + z + 1)-1=(x + 1)[y(z + 1)+(z + 1)]-1=(x + 1)(y + 1)(z + 1)-1=2012$，$\therefore (x + 1)(y + 1)(z + 1)=2013$，$\because x$，$y$，$z$都大于$0$，①当$x + 1$，$y + 1$，$z + 1$中没有一个等于$1$时，$\therefore 2013 = 3×11×61$，共有$6$种；②当$x + 1$，$y + 1$，$z + 1$中有一个等于$1$时，I.当$x + 1 = 1$时，即$x = 0$时，$\therefore 2013 = 3×671 = 11×183 = 61×33$，有$6$种，II.当$y + 1 = 1$时，即$y = 0$时，$\therefore 2013 = 3×671 = 11×183 = 61×33$，有$6$种，III.当$z + 1 = 1$时，即$z = 0$时，$\therefore 2013 = 3×671 = 11×183 = 61×33$，有$6$种，③当$x + 1$，$y + 1$，$z + 1$中有两个等于$1$时，共$3$种，$4×6 + 3 = 27$（种）.故原方程的非负整数解有$27$组.