《2025年全品学练考九年级数学上册人教版》

4. 下面是小聪同学用配方法解方程$2x^2 + 4x - 1 = 0$的过程,请仔细阅读后,解答下面的问题.
解:移项,得$2x^2 + 4x = 1$.①
二次项系数化为1,得$x^2 + 2x = \frac{1}{2}$.②
配方,得$x^2 + 2x + 1 = \frac{1}{2}$,即$(x + 1)^2 = \frac{1}{2}$.③
由此可得$x + 1 = \pm \frac{\sqrt{2}}{2}$,④
$x_1 = -1 + \frac{\sqrt{2}}{2},x_2 = -1 - \frac{\sqrt{2}}{2}$.⑤
(1)小聪的解答过程是从第③步开始出现错误的,错误的原因是配方时等号右边没有加上 1;
(2)用这种方法解方程:$2x^2 - 4x - 3 = 0$.
解：移项，得$2x^2 - 4x = 3$.
二次项系数化为1，得$x^2 - 2x = \frac{3}{2}$.
配方，得$x^2 - 2x + 1 = \frac{3}{2} + 1$，即$(x - 1)^2 = \frac{5}{2}$.
由此可得$x - 1 = \pm \frac{\sqrt{10}}{2}$，
$x_1 = 1 + \frac{\sqrt{10}}{2}, x_2 = 1 - \frac{\sqrt{10}}{2}$.

答案：
(1)③ 配方时等号右边没有加上 1
(2)$x_{1}=1+\frac {\sqrt {10}}{2},x_{2}=1-\frac {\sqrt {10}}{2}$

5. 解下列方程:
(1)$2x^2 - 3x + 1 = 0$;$x_{1}=1,x_{2}=\frac {1}{2}$
(2)$2x(x + \sqrt{2}) - 1 = 0$;$x_{1}=\frac {-\sqrt {2}+2}{2},x_{2}=\frac {-\sqrt {2}-2}{2}$
(3)$4x^2 - 3x - 5 = x - 2$.$x_{1}=\frac {3}{2},x_{2}=-\frac {1}{2}$

答案：
(1)$x_{1}=1,x_{2}=\frac {1}{2}$
(2)$x_{1}=\frac {-\sqrt {2}+2}{2},x_{2}=\frac {-\sqrt {2}-2}{2}$
(3)$x_{1}=\frac {3}{2},x_{2}=-\frac {1}{2}$

6. 阅读材料,解答问题.
解方程:$(4x - 1)^2 - 10(4x - 1) + 24 = 0$.
解:把$4x - 1$视为一个整体,设$4x - 1 = y$,
则原方程可化为$y^2 - 10y + 24 = 0$,
解得$y_1 = 6,y_2 = 4$.
$\therefore 4x - 1 = 6$,或$4x - 1 = 4$,
$\therefore x_1 = \frac{7}{4},x_2 = \frac{5}{4}$.
以上方法就叫换元法,达到简化或降次的目的,体现了转化的思想.
请利用换元法解下列方程:
(1)$(3x - 5)^2 + 4(3x - 5) + 3 = 0$;
(2)$x^4 - x^2 - 6 = 0$.

答案：
(1)$x_{1}=\frac {2}{3},x_{2}=\frac {4}{3}$
(2)$x_{1}=\sqrt {3},x_{2}=-\sqrt {3}$