《2025年实验班提优训练九年级数学上册人教版》

12. 某中学数学兴趣小组对关于x的方程$(m+1)x^{m^{2}+1}+(m-2)x-1= 0$提出了下列问题:
(1)是否存在m的值,使方程为一元二次方程? 若存在,求出m的值,并写出此方程.
(2)是否存在m的值,使方程为一元一次方程? 若存在,求出m的值,并解此方程.

答案：
(1)存在.根据一元二次方程的定义可得$m^{2}+1=2$,且$m+1≠0$,解得$m=1$,此时方程为$2x^{2}-x-1=0$.
(2)存在.由题意,得$m^{2}+1=1$或$m+1=0$时,原方程可能为一元一次方程.当$m^{2}+1=1$时,$m=0$,此时方程为$-x-1=0$,解得$x=-1$;当$m+1=0$时,$m=-1$,此时方程为$-3x-1=0$,解得$x=-\frac {1}{3}$.

13. (2024·扬州邗江区梅岭中学模拟改编)关于x的一元二次方程$a_{1}(x-m)^{2}+k= 0与a_{2}(x-m)^{2}+k= 0$称为“同族二次方程”,如$2(x-3)^{2}+4= 0与3(x-3)^{2}+4= 0$是“同族二次方程”. 现有关于x的一元二次方程$2(x-1)^{2}+1= 0与(a+2)x^{2}+(b-4)x+8= 0$是“同族二次方程”,求代数式ab的值.-50

答案： $\because 2(x-1)^{2}+1=0$与$(a+2)x^{2}+(b-4)x+8=0$是“同族二次方程”，$\therefore (a+2)x^{2}+(b-4)x+8=(a+2)(x-1)^{2}+1$,即$(a+2)x^{2}+(b-4)x+8=(a+2)x^{2}-2(a+2)\cdot x+a+3$,$\therefore \left\{\begin{array}{l} -2(a+2)=b-4,\\ a+3=8,\end{array}\right. $解得$\left\{\begin{array}{l} a=5,\\ b=-10,\end{array}\right. $$\therefore ab=-50$.

14. 分类讨论思想 若$x^{a}-3x^{a-b}+1= 0$是关于x的一元二次方程,求a,b的值.
下面是两位同学的解法:
甲:根据题意,得$\left\{\begin{array}{l} a= 2,\\ a-b= 1,\end{array}\right. 解得\left\{\begin{array}{l} a= 2,\\ b= 1.\end{array}\right. $
乙:根据题意,得$\left\{\begin{array}{l} a= 2,\\ a-b= 1\end{array}\right. 或\left\{\begin{array}{l} a= 1,\\ a-b= 2,\end{array}\right. $
解得$\left\{\begin{array}{l} a= 2,\\ b= 1\end{array}\right. 或\left\{\begin{array}{l} a= 1,\\ b= -1.\end{array}\right. $
你认为上述两位同学的解答是否正确,为什么? 如果不对,请给出正确的答案.
上述两位同学的解法都不正确.理由如下：$\because x^{a}-3x^{a-b}+1=0$是关于x的一元二次方程,∴分以下几种情况：
一元二次方程可以缺少一次项
①$\left\{\begin{array}{l} a=2,\\ a-b=0,\end{array}\right. $解得$\left\{\begin{array}{l} a=2,\\ b=2;\end{array}\right. $
②$\left\{\begin{array}{l} a=2,\\ a-b=1,\end{array}\right. $解得$\left\{\begin{array}{l} a=2,\\ b=1;\end{array}\right. $
③$\left\{\begin{array}{l} a=2,\\ a-b=2,\end{array}\right. $解得$\left\{\begin{array}{l} a=2,\\ b=0;\end{array}\right. $
④$\left\{\begin{array}{l} a=0,\\ a-b=2,\end{array}\right. $解得$\left\{\begin{array}{l} a=0,\\ b=-2;\end{array}\right. $
⑤$\left\{\begin{array}{l} a=1,\\ a-b=2,\end{array}\right. $解得$\left\{\begin{array}{l} a=1,\\ b=-1.\end{array}\right. $
综上所述,$\left\{\begin{array}{l} a=2,\\ b=2\end{array}\right. $或$\left\{\begin{array}{l} a=2,\\ b=1\end{array}\right. $或$\left\{\begin{array}{l} a=2,\\ b=0\end{array}\right. $或$\left\{\begin{array}{l} a=0,\\ b=-2\end{array}\right. $或$\left\{\begin{array}{l} a=1,\\ b=-1.\end{array}\right. $

答案： 上述两位同学的解法都不正确.理由如下：$\because x^{a}-3x^{a-b}+1=0$是关于x的一元二次方程,
∴分以下几种情况：
一元二次方程可以缺少一次项
①$\left\{\begin{array}{l} a=2,\\ a-b=0,\end{array}\right. $解得$\left\{\begin{array}{l} a=2,\\ b=2;\end{array}\right. $
②$\left\{\begin{array}{l} a=2,\\ a-b=1,\end{array}\right. $解得$\left\{\begin{array}{l} a=2,\\ b=1;\end{array}\right. $
③$\left\{\begin{array}{l} a=2,\\ a-b=2,\end{array}\right. $解得$\left\{\begin{array}{l} a=2,\\ b=0;\end{array}\right. $
④$\left\{\begin{array}{l} a=0,\\ a-b=2,\end{array}\right. $解得$\left\{\begin{array}{l} a=0,\\ b=-2;\end{array}\right. $
⑤$\left\{\begin{array}{l} a=1,\\ a-b=2,\end{array}\right. $解得$\left\{\begin{array}{l} a=1,\\ b=-1.\end{array}\right. $
综上所述,$\left\{\begin{array}{l} a=2,\\ b=2\end{array}\right. $或$\left\{\begin{array}{l} a=2,\\ b=1\end{array}\right. $或$\left\{\begin{array}{l} a=2,\\ b=0\end{array}\right. $或$\left\{\begin{array}{l} a=0,\\ b=-2\end{array}\right. $或$\left\{\begin{array}{l} a=1,\\ b=-1.\end{array}\right. $
素养考向 本题是一道探究性问题,考查学生对一元二次方程概念的理解和运用,也考查了学生对于分类讨论数学思想的掌握情况.

15. (2024·西宁中考)如图,小区物业规划在一个长60m,宽22m的矩形场地ABCD上,修建一个小型停车场,阴影部分为停车位所在区域,两侧是宽xm的道路,中间是宽2xm的道路. 如果阴影部分的总面积是$600m^{2}$,那么x满足的方程是(A).

A. $x^{2}-41x+180= 0$
B. $x^{2}-41x+225= 0$
C. $x^{2}-41x+30= 0$
D. $x^{2}-41x-270= 0$

答案： A [解析]
∵矩形场地ABCD的长为60m,宽为22m,且所修建停车位的两侧是宽xm的道路,中间是宽2xm的道路,
∴停车位(即阴影部分)可合成长为$(60-2x)m$,宽为$(22-2x)m$的矩形.根据题意,得$(60-2x)\cdot (22-2x)=600$,化简,得$x^{2}-41x+180=0$.故选A.