登录
期末提优
答案
1. 已知关于 $ x $ 的一元二次方程 $(m-1)x^2 - 2x + m^2 - m = 0$ 有一根为 0,则 $ m $ 的值是（ ）
A. 0
B. 1
C. 0 或 1
D. 0 或 -1
A. 0
B. 1
C. 0 或 1
D. 0 或 -1
答案
A
2. (2024·泰州期末)若$m$是方程$x^2 - 3x - 2 = 0$的一个根，则$\dfrac{m+1}{2} · (m-3) ÷ \dfrac{m^2 - 1}{m^2 - m}$的值为（ ）
A. $-1$
B. $1$
C. $\pm1$
D. $0$
A. $-1$
B. $1$
C. $\pm1$
D. $0$
答案
B
3. (2024·成都中考)若m、n是一元二次方程$x^2 - 5x + 2 = 0$的两个实数根，则$m + (n - 2)^2$的值为______．
答案
7
4. (2024·南通月考)关于$x$的方程$a(x+m)^2+b=0$的解是$x_1=-3,x_2=2(a、b、m$为常数，$a≠0)$，则方程$a(2x+m+1)^2+b=0$的解是______．
答案
$x_1 = -2,x_2=\frac{1}{2}$
5. (2024·苏州期中)用适当的方法解方程:
(1)$3x(x-1)=2(x-1)$;
(2)$(x+1)(x-1)=2\sqrt{2}x$.
(1)$3x(x-1)=2(x-1)$;
(2)$(x+1)(x-1)=2\sqrt{2}x$.
答案
解： $\begin{aligned}3x(x - 1)&=2(x - 1)\\3x(x - 1)-2(x - 1)&=0\\(x - 1)(3x - 2)&=0\end{aligned}$ 则$x - 1 = 0$或$3x - 2 = 0，$
解得$x_1 = 1，$$x_2=\frac{2}{3}。$ ; 解：$ \begin {aligned}(x + 1)(x - 1)&=2\sqrt {2}x\\x ^2-1&=2\sqrt {2}x\\x ^2-2\sqrt {2}x - 1&=0\end {aligned}$$ $这里$a = 1，$$b=-2\sqrt {2}，$$c = - 1，$$∆=b^2-4ac=(-2\sqrt {2})^2-4×1×(-1)$$=8 + 4 = 12。$$ x=\frac {-b\pm \sqrt {∆}}{2a}=\frac {2\sqrt {2}\pm \sqrt {12}}{2}$$=\frac {2\sqrt {2}\pm 2\sqrt {3}}{2}=\sqrt {2}\pm \sqrt {3}，$所以$x_{1}=\sqrt {2}+\sqrt {3}，$$x_{2}=\sqrt {2}-\sqrt {3}。$
解得$x_1 = 1，$$x_2=\frac{2}{3}。$ ; 解：$ \begin {aligned}(x + 1)(x - 1)&=2\sqrt {2}x\\x ^2-1&=2\sqrt {2}x\\x ^2-2\sqrt {2}x - 1&=0\end {aligned}$$ $这里$a = 1，$$b=-2\sqrt {2}，$$c = - 1，$$∆=b^2-4ac=(-2\sqrt {2})^2-4×1×(-1)$$=8 + 4 = 12。$$ x=\frac {-b\pm \sqrt {∆}}{2a}=\frac {2\sqrt {2}\pm \sqrt {12}}{2}$$=\frac {2\sqrt {2}\pm 2\sqrt {3}}{2}=\sqrt {2}\pm \sqrt {3}，$所以$x_{1}=\sqrt {2}+\sqrt {3}，$$x_{2}=\sqrt {2}-\sqrt {3}。$
6. 解方程：
(1) (2023·凉山州中考) $\dfrac{x}{x+1} = \dfrac{2}{x^2 - 1}$．
(2) (2024·上海中考) $\begin{cases} x^2 - 3xy - 4y^2 = 0, \\ x + 2y = 6. \end{cases}$
(1) (2023·凉山州中考) $\dfrac{x}{x+1} = \dfrac{2}{x^2 - 1}$．
(2) (2024·上海中考) $\begin{cases} x^2 - 3xy - 4y^2 = 0, \\ x + 2y = 6. \end{cases}$
答案
解： $\begin{aligned}\frac{x}{x + 1}&=\frac{2}{x^{2}-1}\\\frac{x}{x + 1}&=\frac{2}{(x + 1)(x - 1)}\\x(x - 1)&=2\\x^{2}-x - 2&=0\\(x + 1)(x - 2)&=0\end{aligned}$ 解得$x_1=-1，$$x_2 = 2。$ 检验：当$x=-1$时，$(x + 1)(x - 1)=0，$
所以$x=-1$是增根；当$x = 2$时，$(x + 1)(x - 1)=3\neq0，$
所以原方程的解为$x = 2。$ ; 解：$ \begin {cases}x^2-3xy - 4y^2=0&①\\x + 2y=6&②\end {cases}$$ $由$②$得$x=6 - 2y，$代入$①$中得$(6 - 2y)^2-3(6 - 2y)y-4y^2=0，$$ $即$(36-24y + 4y^2)-18y+6y^2-4y^2=0，$$6y^2-42y + 36=0，$$6(y^2-7y + 6)=0，$$ $解得$y = 1$或$y = 6。$$ $当$y = 1$时，$x=6-2×1 = 4；$当$y = 6$时，$x=6-2×6=-6。$$ $所以方程组的解为$\begin {cases}x = 4\\y = 1\end {cases}$或$\begin {cases}x=-6\\y = 6\end {cases}。$
所以$x=-1$是增根；当$x = 2$时，$(x + 1)(x - 1)=3\neq0，$
所以原方程的解为$x = 2。$ ; 解：$ \begin {cases}x^2-3xy - 4y^2=0&①\\x + 2y=6&②\end {cases}$$ $由$②$得$x=6 - 2y，$代入$①$中得$(6 - 2y)^2-3(6 - 2y)y-4y^2=0，$$ $即$(36-24y + 4y^2)-18y+6y^2-4y^2=0，$$6y^2-42y + 36=0，$$6(y^2-7y + 6)=0，$$ $解得$y = 1$或$y = 6。$$ $当$y = 1$时，$x=6-2×1 = 4；$当$y = 6$时，$x=6-2×6=-6。$$ $所以方程组的解为$\begin {cases}x = 4\\y = 1\end {cases}$或$\begin {cases}x=-6\\y = 6\end {cases}。$
7. (2025·无锡期中)我们已经学习了一元二次方程的多种解法,其基本思路是将二次方程通过“降次”转化为一次方程求解.按照同样的思路,我们可以将更高次的方程“降次”,转化为二次方程或一次方程进行求解.
①换元法求解四次方程:$x^4 - 5x^2 + 4 = 0$.
设$x^2 = y$,则原方程可变为$y^2 - 5y + 4 = 0$,解得$y_1 = 1,y_2 = 4$,当$y = 1$时,即$x^2 = 1,\therefore x = \pm1$;当$y = 4$时,即$x^2 = 4,\therefore x = \pm2,\therefore$原方程有四个根:$x_1 = 1,x_2 = -1,x_3 = 2,x_4 = -2$.
②因式分解法求解三次方程:$x^3 - 5x + 2 = 0$,将其变形为$x^3 - (4 + 1)x + 2 = 0,\therefore x^3 - 4x - x + 2 = 0$,
$\therefore (x^3 - 4x) - (x - 2) = 0,\therefore x(x + 2)(x - 2) - (x - 2) = 0,\therefore (x - 2)(x^2 + 2x - 1) = 0,\therefore x - 2 = 0$或$x^2 + 2x - 1 = 0,\therefore$原方程有三个根:$x_1 = 2,x_2 = -1+\sqrt{2},x_3 = -1-\sqrt{2}$.
(1)仿照以上方法解方程:
①$x^4 + x^2 - 12 = 0$;
②$x^3 - 17x + 4 = 0$;
(2)已知:$x^2 - x - 1 = 0$,且$x＞0$,则$x^4 - 2x^3 + 3x$的值为______.
①换元法求解四次方程:$x^4 - 5x^2 + 4 = 0$.
设$x^2 = y$,则原方程可变为$y^2 - 5y + 4 = 0$,解得$y_1 = 1,y_2 = 4$,当$y = 1$时,即$x^2 = 1,\therefore x = \pm1$;当$y = 4$时,即$x^2 = 4,\therefore x = \pm2,\therefore$原方程有四个根:$x_1 = 1,x_2 = -1,x_3 = 2,x_4 = -2$.
②因式分解法求解三次方程:$x^3 - 5x + 2 = 0$,将其变形为$x^3 - (4 + 1)x + 2 = 0,\therefore x^3 - 4x - x + 2 = 0$,
$\therefore (x^3 - 4x) - (x - 2) = 0,\therefore x(x + 2)(x - 2) - (x - 2) = 0,\therefore (x - 2)(x^2 + 2x - 1) = 0,\therefore x - 2 = 0$或$x^2 + 2x - 1 = 0,\therefore$原方程有三个根:$x_1 = 2,x_2 = -1+\sqrt{2},x_3 = -1-\sqrt{2}$.
(1)仿照以上方法解方程:
①$x^4 + x^2 - 12 = 0$;
②$x^3 - 17x + 4 = 0$;
(2)已知:$x^2 - x - 1 = 0$,且$x＞0$,则$x^4 - 2x^3 + 3x$的值为______.
答案
$1+\sqrt{5}$ ; 解$：(1)①$令$t=x²(t≥0)$则$t²+t-12=0$∴$(t-3)(t+4)=0$∴$t=3$或$t=-4($舍$)$∴$x²=3$∴$x=±\sqrt {3}$$②$将$x³-17x+4=0$变形为$x³-(16+1)x+4=0$$(x³-16x)+(4-x)=0$$x(x²-16)+(4-x)=0$$x(x-4)(x+4)-(x-4)=0$$(x²+4x-1)(x-4)=0$∴$x_{1}=4，x_{2}=-2+\sqrt {5}，x_{3}=-2-\sqrt {5}$
