《2025年实验班提优训练九年级数学上册人教版》

【例1】(“大梦杯”福建初中数学竞赛)若关于x的方程$4x^{2}+4mx-3m-1= 0$有两个相等的实数根,则$m^{3}+4m^{2}+4m-2$的值为(A).
A. -3 B. -2 C. -1 D. 1

答案： 解析:依题意,得$Δ=(4m)^{2}-4×4×(-3m-1)= 16m^{2}+16(3m+1)= 0,$
即$m^{2}+3m+1= 0,$
$\therefore m^{2}= -3m-1,m^{2}+3m= -1,$
$\therefore m^{3}+4m^{2}+4m-2= m(-3m-1)+4m^{2}+4m-2= m^{2}+3m-2= -1-2= -3.$
答案:A

【例2】[全国初中数学竞赛(甘肃赛区)初赛]方程$(x^{2}+x-1)^{x+3}= 1$的所有整数解的个数是(B).
A. 5 B. 4 C. 3 D. 2

答案： 解析:①当$x+3= 0,x^{2}+x-1≠0$时,解得$x= -3;$
②当$x^{2}+x-1= 1$时,解得$x= -2$或1;
③当$x^{2}+x-1= -1,x+3$为偶数时,解得$x= -1.$
因此原方程的所有整数解是-3,-2,1,-1,共4个.
答案:B

1.(“大梦杯”福建初中数学竞赛)满足方程$x^{2}-4xy+19y^{2}= 151的整数对(x,y)$有(C).
A. 0对
B. 2对
C. 4对
D. 6对

答案： C [解析]方程$x^{2}-4xy+19y^{2}=151$可化为$(x-2y)^{2}=151-15y^{2}$.
依题意,得$A=151-15y^{2}$为完全平方数.
由$A=151-15y^{2}≥0$,得$y^{2}≤\frac {151}{15}$.
结合$y$为整数,得$y^{2}≤10.\therefore y^{2}=0,1,4,9$.
当$y^{2}=0$时,$A=151-15y^{2}=151$,不是完全平方数;当$y^{2}=1$时,$A=151-15y^{2}=136$,不是完全平方数;当$y^{2}=4$时,$A=151-15y^{2}=91$,不是完全平方数;当$y^{2}=9$时,$A=151-15y^{2}=16=4^{2}$.
$\therefore$原方程可化为$\left\{\begin{array}{l} y^{2}=9\\ (x-2y)^{2}=16\end{array}\right.$.
$\therefore \left\{\begin{array}{l} y=3\\ (x-6)^{2}=16\end{array}\right.$或$\left\{\begin{array}{l} y=-3\\ (x+6)^{2}=16\end{array}\right.$.
$\therefore \left\{\begin{array}{l} y=3\\ x-6=4\end{array}\right.$或$\left\{\begin{array}{l} y=3\\ x-6=-4\end{array}\right.$或$\left\{\begin{array}{l} y=-3\\ x+6=4\end{array}\right.$或$\left\{\begin{array}{l} y=-3\\ x+6=-4\end{array}\right.$.
$\therefore \left\{\begin{array}{l} x=10\\ y=3\end{array}\right.$或$\left\{\begin{array}{l} x=2\\ y=3\end{array}\right.$或$\left\{\begin{array}{l} x=-2\\ y=-3\end{array}\right.$或$\left\{\begin{array}{l} x=-10\\ y=-3\end{array}\right.$.
$\therefore$满足方程$x^{2}-4xy+19y^{2}=151$的整数对$(x,y)$为$(10,3),(2,3),(-2,-3)$或$(-10,-3)$,共有4对.故选C.

2.[全国初中数学竞赛(甘肃赛区)初赛]设$x_{1},x_{2}是方程x^{2}-2(k+1)x+k^{2}+2= 0$的两个实数根,且$(x_{1}+1)(x_{2}+1)= 8$,则k的值是____1____.

答案： 1 [解析]由题意,得$\Delta =[-2(k+1)]^{2}-4(k^{2}+2)≥0$,解得$k≥\frac {1}{2}.\because x_{1}+x_{2}=2(k+1),x_{1}x_{2}=k^{2}+2$.
$\therefore (x_{1}+1)(x_{2}+1)=x_{1}x_{2}+(x_{1}+x_{2})+1=k^{2}+2+2(k+1)+1=k^{2}+2k+5=8$.
$\therefore k^{2}+2k-3=0$,解得$k=-3$(舍去)或$k=1$.

3.[全国初中数学竞赛(天津赛区)复赛]实数a,b,c,d满足:一元二次方程$x^{2}+cx+d= 0$的两根为a,b,一元二次方程$x^{2}+ax+b= 0$的两根为c,d,则所有满足条件的数组$(a,b,c,d)$为$(1,-2,1,-2)$或$(a,0,-a,0)$($a$为任意实数).

答案： $(1,-2,1,-2)$或$(a,0,-a,0)$($a$为任意实数) [解析]$\because$实数$a,b,c,d$满足:一元二次方程$x^{2}+cx+d=0$的两根为$a,b$,一元二次方程$x^{2}+ax+b=0$的两根为$c,d$.
$\therefore \left\{\begin{array}{l} a+b=-c\\ ab=d\\ c+d=-a\\ cd=b\end{array}\right.$,解得$\left\{\begin{array}{l} a=1\\ b=-2\\ c=1\\ d=-2\end{array}\right.$或$\left\{\begin{array}{l} a=-c\\ b=0\\ d=0\end{array}\right.$($a$为任意实数).
故答案为$(1,-2,1,-2)$或$(a,0,-a,0)$($a$为任意实数).

4.[全国初中数学竞赛(孝感赛区)初赛改编]如果关于x的方程$x^{2}+kx+\frac {3}{4}k^{2}-3k+\frac {9}{2}= 0的两个实数根分别为x_{1},x_{2}$,那么$\frac {x_{1}^{2024}}{x_{2}^{2025}}$的值为$-\frac {2}{3}$.

答案： $-\frac {2}{3}$ [解析]$\because$方程有实数根,$\therefore \Delta =b^{2}-4ac≥0$,即$k^{2}-4(\frac {3}{4}k^{2}-3k+\frac {9}{2})≥0.\therefore -2(k-3)^{2}≥0$.
$\because (k-3)^{2}≥0,\therefore k-3=0$,即$k=3,\therefore$原方程为$x^{2}+3x+\frac {9}{4}=0,\therefore x_{1}=x_{2}=-\frac {3}{2}.\therefore \frac {x_{1}^{2024}}{x_{2}^{2025}}=(\frac {x_{1}}{x_{2}})^{2024}\cdot \frac {1}{x_{2}}=\frac {1}{x_{2}}=-\frac {2}{3}$.

5.[全国初中数学竞赛(河南赛区)预赛]已知α,β是方程$x^{2}+2x-1= 0$的两根,则$α^{3}+5β+10$的值为-2.

答案： -2 [解析]$\because \alpha$是方程$x^{2}+2x-1=0$的根,$\therefore \alpha ^{2}=1-2\alpha .\therefore \alpha ^{3}=\alpha ^{2}\cdot \alpha =(1-2\alpha )\cdot \alpha =\alpha -2\alpha ^{2}=\alpha -2(1-2\alpha )=5\alpha -2$.又$\alpha +\beta =-2,\therefore \alpha ^{3}+5\beta +10=(5\alpha -2)+5\beta +10=5(\alpha +\beta )+8=5×(-2)+8=-2$.

6.[全国初中数学竞赛(浙江赛区)复赛]设整数a使得关于x的一元二次方程$5x^{2}-5ax+26a-143= 0$的两个根都是整数,则a的值是____18____.

答案： 18 [解析]$5x^{2}-5ax+26a-143=0$可化为$25x^{2}-25ax+(130a-26^{2})-39=0$,即$(5x-26)(5x-5a+26)=39.\because x,a$都是整数,故$(5x-26),(5x-5a+26)$都分别为整数,而只存在$39=1×39$或$39×1$或$3×13$或$13×3$或$(-1)×(-39)$或$(-39)×(-1)$或$(-3)×(-13)$或$(-13)×(-3)$八种情况,①当$5x-26=1,5x-5a+26=39$,相减解得$a=2.8$不符合;②当$5x-26=39,5x-5a+26=1$,相减解得$a=18$;③当$5x-26=3,5x-5a+26=13$,相减解得$a=8.4$不符合;④当$5x-26=13,5x-5a+26=3$,相减解得$a=12.4$不符合;⑤当$5x-26=-1,5x-5a+26=-39$,相减解得$a=18$;⑥当$5x-26=-39,5x-5a+26=-1$,相减解得$a=2.8$不符合;⑦当$5x-26=-3,5x-5a+26=-13$,相减解得$a=12.4$不符合;⑧当$5x-26=-13,5x-5a+26=-3$,相减解得$a=8.4$不符合.$\therefore$当$a=18$时,方程为$5x^{2}-90x+325=0$,两根为13,5.故答案为18.

7.(“大梦杯”福建初中数学竞赛)已知关于x的方程$x^{2}-kx-k+9999= 0$的两根都是素数,求k的值.

答案： 设方程$x^{2}-kx-k+9999=0$的两根分别为$p,q$,
则由一元二次方程根与系数的关系,知$\left\{\begin{array}{l} p+q=k\\ pq=-k+9999\end{array}\right.$.
$\therefore pq+p+q=9999$.
$\therefore (p+1)(q+1)=10000=2^{4}×5^{4}$.
显然$p,q$都不等于2,因此$p,q$都是奇数.
$\therefore \frac {p+1}{2}\cdot \frac {q+1}{2}=2^{2}×5^{4}$.
①若$\frac {p+1}{2},\frac {q+1}{2}$中有一个数为奇数,不妨设$\frac {p+1}{2}$为奇数,
则$\frac {p+1}{2}=5^{m}$,其中$m=1,2,3,4$.
当$m=1$时,$p=9$,不是素数,舍去;
当$m=2$时,$p=49$,不是素数,舍去;
当$m=3$时,$p=249$,不是素数,舍去;
当$m=4$时,$p=1249$,是素数,此时$\frac {q+1}{2}=2^{2}$,解得$q=7$,也是素数.
$\therefore p=1249,q=7,k=p+q=1256$,符合要求.
②若$\frac {p+1}{2},\frac {q+1}{2}$都是偶数,则$\frac {p+1}{4}\cdot \frac {q+1}{4}=5^{4}$.
不妨设$p≤q$,则当$\frac {p+1}{4}=5^{1},\frac {q+1}{4}=5^{3}$时,$p=3,q=2499,q$不是素数,舍去;当$\frac {p+1}{4}=5^{1},\frac {q+1}{4}=5^{3}$时,$p=19,q=499,p,q$都是素数;当$\frac {p+1}{4}=5^{2},\frac {q+1}{4}=5^{2}$时,$p=99,q=99,p,q$都不是素数,舍去.
$\therefore p=19,q=499,k=p+q=518$,符合要求.
综上所述,$k=518$或1256.

8.[“数学周报杯”全国初中数学竞赛(天津赛区)复赛]已知关于x的一元二次方程$x^{2}+cx+a= 0$的两个整数根恰好比方程$x^{2}+ax+b= 0$的两个根都大1,求$a+b+c$的值.-3或29

答案： 设方程$x^{2}+ax+b=0$的两个根为$\alpha ,\beta$,
由题可知$\alpha ,\beta$为整数,不妨设$\alpha ≤\beta$,
则方程$x^{2}+cx+a=0$的两根为$\alpha +1,\beta +1$,
$\therefore \alpha +\beta =-a,(\alpha +1)(\beta +1)=a$.
两式相加,得$\alpha \beta +2\alpha +2\beta +1=0$,即$(\alpha +2)(\beta +2)=3$.
$\therefore \left\{\begin{array}{l} \alpha +2=1\\ \beta +2=3\end{array}\right.$或$\left\{\begin{array}{l} \alpha +2=-3\\ \beta +2=-1\end{array}\right.$,解得$\left\{\begin{array}{l} \alpha =-1\\ \beta =1\end{array}\right.$或$\left\{\begin{array}{l} \alpha =-5\\ \beta =-3\end{array}\right.$.
$\therefore a=-(\alpha +\beta )=-[(-1)+1]=0,b=\alpha \beta =-1×1=-1,c=-[(\alpha +1)+(\beta +1)]=-[(-1+1)+(1+1)]=-2$,
或$a=-(\alpha +\beta )=-[(-5)+(-3)]=8,b=\alpha \beta =(-5)×(-3)=15,c=-[(\alpha +1)+(\beta +1)]=-[(-5+1)+(-3+1)]=6$,
$\therefore a=0,b=-1,c=-2$或$a=8,b=15,c=6$,
$\therefore a+b+c=0+(-1)+(-2)=-3$或$a+b+c=8+15+6=29$.
故$a+b+c=-3$或29.