《2025年高效精练九年级数学上册苏科版》

12. (2024·绥化)小影与小冬一起写作业,在解一道一元二次方程时,小影在化简过程中写错了常数项,因而得到方程的两个根是6和1;小冬在化简过程中写错了一次项的系数,因而得到方程的两个根是-2和-5.则原来的方程是 (B)
A. $x^{2}+6x+5=0$
B. $x^{2}-7x+10=0$
C. $x^{2}-5x+2=0$
D. $x^{2}-6x-10=0$

答案： B

13. (2023·乐山)若关于x的一元二次方程$x^{2}-8x+m=0$两根为$x_{1},x_{2}$,且$x_{1}=3x_{2}$,则m的值为 (C)
A. 4
B. 8
C. 12
D. 16

答案： C

14. 若实数a,b分别满足$a^{2}-4a+3=0,b^{2}-4b+3=0$,且$a≠b$,则$\frac {1}{a}+\frac {1}{b}$的值为$\frac {4}{3}$.

答案： $\frac {4}{3}$

15. 已知关于x的一元二次方程$x^{2}-4x-2m+5=0$有两个不相等的实数根.
(1)求实数m的取值范围;
(2)若该方程的两个根都是符号相同的整数,求整数m的值.

答案：
(1) 根据题意得$\Delta = (-4)^{2} - 4(-2m + 5) ＞ 0$，解得$m ＞ \frac {1}{2}$；所以实数 m 的取值范围为$m ＞ \frac {1}{2}$.
(2) 设$x_{1}$，$x_{2}$是方程的两根，根据题意得$x_{1} + x_{2} = 4 ＞ 0$，$x_{1}x_{2} = -2m + 5 ＞ 0$，解得$m ＜ \frac {5}{2}$，而$m ＞ \frac {1}{2}$，所以 m 的取值范围为$\frac {1}{2} ＜ m ＜ \frac {5}{2}$，因为 m 为整数，所以$m = 1$或$m = 2$，当$m = 1$时，方程两根都是整数；当$m = 2$时，方程两根都不是整数；所以整数 m 的值为 1.

16. (2024·遂宁)已知关于x的一元二次方程$x^{2}-(m+2)x+m-1=0$.
(1)求证:无论m取何值,方程都有两个不相等的实数根;
(2)如果方程的两个实数根为$x_{1},x_{2}$,且$x_{1}^{2}+x_{2}^{2}-x_{1}x_{2}=9$,求m的值.-2或1

答案：
(1) $x^{2} - (m + 2)x + m - 1 = 0$，$a = 1$，$b = -(m + 2)$，$c = m - 1$，$\Delta = b^{2} - 4ac = [-(m + 2)]^{2} - 4×1×(m - 1) = m^{2} + 4m + 4 - 4m + 4 = m^{2} + 8$. $\because m^{2} ≥ 0$，$\therefore \Delta ＞ 0$. $\therefore$ 无论 m 取何值，方程都有两个不相等的实数根；
(2) 设方程$x^{2} - (m + 2)x + m - 1 = 0$的两个实数根为$x_{1}$，$x_{2}$，则$x_{1} + x_{2} = m + 2$，$x_{1}x_{2} = m - 1$. $\because x_{1}^{2} + x_{2}^{2} - x_{1}x_{2} = 9$，即$(x_{1} + x_{2})^{2} - 3x_{1}x_{2} = 9$，$\therefore (m + 2)^{2} - 3(m - 1) = 9$. 整理，得$m^{2} + m - 2 = 0$，$\therefore (m + 2)(m - 1) = 0$. 解得$m_{1} = -2$，$m_{2} = 1$，$\therefore$ m 的值为 -2 或 1.

17. (2023·通辽)阅读材料:
材料1:关于x的一元二次方程$ax^{2}+bx+c=0(a≠0)$的两个实数根$x_{1},x_{2}$和系数a,b,c,有如下关系:$x_{1}+x_{2}=-\frac {b}{a},x_{1}x_{2}=\frac {c}{a}$.
材料2:已知一元二次方程$x^{2}-x-1=0$的两个实数根分别为m,n,求$m^{2}n+mn^{2}$的值.
解:$\because m,n$是一元二次方程$x^{2}-x-1=0$的两个实数根,
$\therefore m+n=1,mn=-1$.
则$m^{2}n+mn^{2}=mn(m+n)=-1×1=-1$.
根据上述材料,结合你所学的知识,完成下列问题:
(1)应用:一元二次方程$2x^{2}+3x-1=0$的两个实数根为$x_{1},x_{2}$,则$x_{1}+x_{2}=$$-\frac {3}{2}$, $x_{1}x_{2}=$$-\frac {1}{2}$.
(2)类比:已知一元二次方程$2x^{2}+3x-1=0$的两个实数根为m,n,求$m^{2}+n^{2}$的值;
解:$\because$ 一元二次方程$2x^{2} + 3x - 1 = 0$的两根分别为 m，n，$\therefore m + n = -\frac {3}{2}$，$mn = -\frac {1}{2}$，$\therefore m^{2} + n^{2} = (m + n)^{2} - 2mn = \frac {9}{4} + 1 = \frac {13}{4}$
(3)提升:已知实数s,t满足$2s^{2}+3s-1=0,2t^{2}+3t-1=0$且$s≠t$,求$\frac {1}{s}-\frac {1}{t}$的值.
解:$\because$ 实数 s，t 满足$2s^{2} + 3s - 1 = 0$，$2t^{2} + 3t - 1 = 0$，且$s ≠ t$，$\therefore$ s，t 是一元二次方程$2x^{2} + 3x - 1 = 0$的两个实数根，$\therefore s + t = -\frac {3}{2}$，$st = -\frac {1}{2}$. $\because (t - s)^{2} = (t + s)^{2} - 4st = (-\frac {3}{2})^{2} - 4×(-\frac {1}{2}) = \frac {17}{4}$，$\therefore t - s = ±\frac {\sqrt {17}}{2}$，$\therefore \frac {1}{s} - \frac {1}{t} = \frac {t - s}{st} = \frac {±\frac {\sqrt {17}}{2}}{-\frac {1}{2}} = ±\sqrt {17}$

答案：
(1) $-\frac {3}{2}$，$-\frac {1}{2}$；
(2) $\because$ 一元二次方程$2x^{2} + 3x - 1 = 0$的两根分别为 m，n，$\therefore m + n = -\frac {3}{2}$，$mn = -\frac {1}{2}$，$\therefore m^{2} + n^{2} = (m + n)^{2} - 2mn = \frac {9}{4} + 1 = \frac {13}{4}$；
(3) $\because$ 实数 s，t 满足$2s^{2} + 3s - 1 = 0$，$2t^{2} + 3t - 1 = 0$，且$s ≠ t$，$\therefore$ s，t 是一元二次方程$2x^{2} + 3x - 1 = 0$的两个实数根，$\therefore s + t = -\frac {3}{2}$，$st = -\frac {1}{2}$. $\because (t - s)^{2} = (t + s)^{2} - 4st = (-\frac {3}{2})^{2} - 4×(-\frac {1}{2}) = \frac {17}{4}$，$\therefore t - s = ±\frac {\sqrt {17}}{2}$，$\therefore \frac {1}{s} - \frac {1}{t} = \frac {t - s}{st} = \frac {±\frac {\sqrt {17}}{2}}{-\frac {1}{2}} = ±\sqrt {17}$.