《2025年资源与评价黑龙江教育出版社九年级数学上册人教版》

2. 解方程$(3x - 2)^2 = 2(3x - 2)$的恰当方法是(D)
A.直接开平方法
B.配方法
C.公式法
D.因式分解法

答案： D

1. 已知方程$x^2 + 2x - 3 = 0的解是x_1 = 1$,$x_2 = -3$,则方程$(2x + 3)^2 + 2(2x + 3) - 3 = 0$的解是(D)
A.$x_1 = 1$,$x_2 = 3$
B.$x_1 = 1$,$x_2 = -3$
C.$x_1 = -1$,$x_2 = 3$
D.$x_1 = -1$,$x_2 = -3$

答案： D

2. 若$a^2 - 3ab - 4b^2 = 0$,则$\frac{a}{b}$的值为4或-1.

答案： 4或-1

1. 一元二次方程$x^2 - x = 0$的根是(C)
A.$x = 1$
B.$x = 0$
C.$x_1 = 0$,$x_2 = 1$
D.$x_1 = 0$,$x_2 = -1$

答案： C

2. 若方程$(x - 8)(5x + 9) = 0$,则$5x + 9$的值是(D)
A.49
B.0
C.$-\frac{9}{5}$
D.49 或 0

答案： D

3. 如果 0 是一元二次方程$2x^2 - 5mx + (m - 2) = 0$的一根,那么另一根为(D)
A.2
B.3
C.4
D.5

答案： D

4. 方程$x(x + 2) = 0$的解为x₁=0,x₂=-2；$(x + 1)(x - 2) = 0$的解为x₁=-1,x₂=2；$(x + 1)^2 = x + 1$的解为x₁=0,x₂=-1.

答案： x₁=0,x₂=-2；x₁=-1,x₂=2；x₁=0,x₂=-1

5. 用因式分解法解下列方程.
(1)$(2x - 3)^2 - (x - 2)^2 = 0$；
(2)$2(t - 1)^2 + t = 1$.

答案： 解:
(1)
∵[(2x-3)+(x-2)][(2x-3)-(x-2)]=0,
∴(3x-5)(x-1)=0,
∴3x-5=0或x-1=0,
∴x₁=$\frac{5}{3}$,x₂=1.
(2)原方程可化为2(t-1)²+t-1=0,
∴(t-1)(2t-1)=0,
∴t₁=1,t₂=$\frac{1}{2}$.

6. (过程性学习)阿进用因式分解法解一元二次方程$5x^2 - 15x = 6 - 2x$时,他的做法如下：
解：方程两边分解因式,得$5x(x - 3) = 2(3 - x)$,·····································································第一步
方程变形为$5x(x - 3) = -2(x - 3)$,·····································································第二步
方程两边同时除以$(x - 3)$,得$5x = -2$,·····································································第三步
系数化为 1,得$x = -\frac{2}{5}$. ··································第四步
(1)阿进的解法是不正确的,从第______步开始出现错误.
(2)请用阿进的方法完成这个题的解题过程.
(1)三
(2)5x²-15x=6-2x,因式分解,得5x(x-3)=2(3-x),整理,得5x(x-3)=-2(x-3),移项,得5x(x-3)+2(x-3)=0,提公因式,得(x-3)(5x+2)=0,即x-3=0或5x+2=0,
∴x₁=3,x₂=-$\frac{2}{5}$.

答案： 解:
(1)三.
(2)5x²-15x=6-2x,因式分解,得5x(x-3)=2(3-x),整理,得5x(x-3)=-2(x-3),移项,得5x(x-3)+2(x-3)=0,提公因式,得(x-3)(5x+2)=0,即x-3=0或5x+2=0,
∴x₁=3,x₂=-$\frac{2}{5}$.