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1 [2026 河北沧州期中,中]关于 x 的方程 $x^2 - 2mx + m^2 = 4$ 的两个根 $x_1, x_2$ 满足 $x_1 = 2x_2 + 3$，且 $x_1 ＞ x_2$，则 $m$ 的值为 (
A.-3
B.1
C.3
D.9
C
)A.-3
B.1
C.3
D.9
答案
C
【解析】$x^2 - 2mx + m^2 = 4,\therefore (x - m)^2 = 4,\therefore x - m = 2$或$x - m = -2$. $\because x_1 ＞ x_2,\therefore x_1 = m + 2,x_2 = m - 2$. 由$x_1 = 2x_2 + 3$得$m + 2 = 2(m - 2) + 3,\therefore m = 3$. 故选 C.
【解析】$x^2 - 2mx + m^2 = 4,\therefore (x - m)^2 = 4,\therefore x - m = 2$或$x - m = -2$. $\because x_1 ＞ x_2,\therefore x_1 = m + 2,x_2 = m - 2$. 由$x_1 = 2x_2 + 3$得$m + 2 = 2(m - 2) + 3,\therefore m = 3$. 故选 C.
2[2026浙江杭州期中,中]已知关于x的方程$(x+a)(x+b)=0$有$M$个解,关于$x$的方程$(ax+1)(bx+1)=0$有$N$个解,其中$a≠b$,则$M$与$N$的关系为 (
A.$M=N-1$或$M=N+1$
B.$M=N-1$或$M=N+2$
C.$M=N$或$M=N+1$
D.$M=N$或$M=N-1$
C
)A.$M=N-1$或$M=N+1$
B.$M=N-1$或$M=N+2$
C.$M=N$或$M=N+1$
D.$M=N$或$M=N-1$
答案
C
【解析】$(x + a)(x + b) = 0,\therefore x + a = 0$或$x + b = 0$,即$x = -a$或$x = -b$. $\because a ≠ b,\therefore M = 2$. 当$a = 0$,$b ≠ 0$时,方程$(ax + 1)(bx + 1) = 0$变为$bx + 1 = 0$,解得$x = -\frac{1}{b}$,此时$N = 1$; 当$a ≠ 0,b = 0$时,方程$(ax + 1)(bx + 1) = 0$变为$ax + 1 = 0$,解得$x = -\frac{1}{a}$,此时$N = 1$; 当$a ≠ 0$且$b ≠ 0$时,方程$(ax + 1)· (bx + 1) = 0,\therefore ax + 1 = 0$或$bx + 1 = 0$,解得$x = -\frac{1}{a}$或$x = -\frac{1}{b}$,此时$N = 2$. 综上,当$a = 0$或$b = 0$时,$M = 2,N = 1$,则$M = N + 1$; 当$a ≠ 0$且$b ≠ 0$时,$M = 2,N = 2$,则$M = N,\therefore M = N$或$M = N + 1$. 故选 C.
【解析】$(x + a)(x + b) = 0,\therefore x + a = 0$或$x + b = 0$,即$x = -a$或$x = -b$. $\because a ≠ b,\therefore M = 2$. 当$a = 0$,$b ≠ 0$时,方程$(ax + 1)(bx + 1) = 0$变为$bx + 1 = 0$,解得$x = -\frac{1}{b}$,此时$N = 1$; 当$a ≠ 0,b = 0$时,方程$(ax + 1)(bx + 1) = 0$变为$ax + 1 = 0$,解得$x = -\frac{1}{a}$,此时$N = 1$; 当$a ≠ 0$且$b ≠ 0$时,方程$(ax + 1)· (bx + 1) = 0,\therefore ax + 1 = 0$或$bx + 1 = 0$,解得$x = -\frac{1}{a}$或$x = -\frac{1}{b}$,此时$N = 2$. 综上,当$a = 0$或$b = 0$时,$M = 2,N = 1$,则$M = N + 1$; 当$a ≠ 0$且$b ≠ 0$时,$M = 2,N = 2$,则$M = N,\therefore M = N$或$M = N + 1$. 故选 C.
3[2025吉林白城质检,中]【阅读与理解】将多项式$x^2+2x-35$分解因式,我们可以按下面的方法分解:
①拆分二次项与常数项:$x^2=x· x,-35=(-5)×(+7)$.
②交叉相乘,验中项(交叉相乘后的结果相加,其结果须等于多项式中的一次项):
$\begin{array}{r}x\ \ +7=7x\\x\ \ -5=-5x\end{array}$
$7x+(-5x)=2x$.
③横向写出两因式:$x^2+2x-35=(x+7)(x-5)$.
我们把这种分解因式的方法称为十字相乘法.
根据乘法原理:若$ab=0$,则$a=0$或$b=0$,方程$x^2+2x-35=0$可以这样求解:方程左边因式分解得$(x+7)(x-5)=0$,所以原方程的解为$x_1=-7,x_2=5$.
【解决问题】
(1)二次项系数为1试用上述方法和原理解下列方程:
①$x^2-10x+21=0$;②$x^2+5x+4=0$;③$x^2-6x-7=0$.
①拆分二次项与常数项:$x^2=x· x,-35=(-5)×(+7)$.
②交叉相乘,验中项(交叉相乘后的结果相加,其结果须等于多项式中的一次项):
$\begin{array}{r}x\ \ +7=7x\\x\ \ -5=-5x\end{array}$
$7x+(-5x)=2x$.
③横向写出两因式:$x^2+2x-35=(x+7)(x-5)$.
我们把这种分解因式的方法称为十字相乘法.
根据乘法原理:若$ab=0$,则$a=0$或$b=0$,方程$x^2+2x-35=0$可以这样求解:方程左边因式分解得$(x+7)(x-5)=0$,所以原方程的解为$x_1=-7,x_2=5$.
【解决问题】
(1)二次项系数为1试用上述方法和原理解下列方程:
①$x^2-10x+21=0$;②$x^2+5x+4=0$;③$x^2-6x-7=0$.
答案
(1)①方程左边因式分解,得$(x - 7)(x - 3) = 0,\therefore x - 7 = 0$或$x - 3 = 0,\therefore x_1 = 7,x_2 = 3$.
②方程左边因式分解,得$(x + 1)(x + 4) = 0,\therefore x + 1 = 0$或$x + 4 = 0,\therefore x_1 = -1,x_2 = -4$.
③方程左边因式分解,得$(x + 1)(x - 7) = 0,\therefore x + 1 = 0$或$x - 7 = 0,\therefore x_1 = -1,x_2 = 7$.
②方程左边因式分解,得$(x + 1)(x + 4) = 0,\therefore x + 1 = 0$或$x + 4 = 0,\therefore x_1 = -1,x_2 = -4$.
③方程左边因式分解,得$(x + 1)(x - 7) = 0,\therefore x + 1 = 0$或$x - 7 = 0,\therefore x_1 = -1,x_2 = 7$.
(2) 二次项系数不为1 解方程:$2024x^2 +2019x -5=0$
答案
(2)方程左边因式分解,得$(2024x - 5)(x + 1) = 0$,$\therefore 2024x - 5 = 0$或$x + 1 = 0$,$\therefore x_1 = \frac{5}{2024}$,$x_2 = -1$.
4[2026广东汕头质检,中]已知关于x的一元二次方程$x^2 - mx + m - 1 = 0$.若平行四边形ABCD的两边AB,AD的长是已知方程的两个实数根.
(1)若平行四边形ABCD是矩形,且$m=5$,求矩形的面积.
(2)当m取何值时,平行四边形ABCD是菱形?求此菱形的边长.
刷素养 走向重高
(1)若平行四边形ABCD是矩形,且$m=5$,求矩形的面积.
(2)当m取何值时,平行四边形ABCD是菱形?求此菱形的边长.
刷素养 走向重高
答案
(1)当$m = 5$时,$x^2 - 5x + 4 = 0$,解得$x_1 = 1$,$x_2 = 4$,即$AB$,$AD$的长分别为 1,4 或 4,1,$\therefore$ 矩形的面积为$1 × 4 = 4$.
(2)$\because$ 平行四边形$ABCD$是菱形,$\therefore AB = AD$,$\therefore \Delta = 0$,$\therefore m^2 - 4(m - 1) = 0$,即$(m - 2)^2 = 0$,解得$m_1 = m_2 = 2$. 当$m = 2$时,方程为$x^2 - 2x + 1 = 0$,解得$x_1 = x_2 = 1$,$\therefore$ 此菱形的边长为 1.
(2)$\because$ 平行四边形$ABCD$是菱形,$\therefore AB = AD$,$\therefore \Delta = 0$,$\therefore m^2 - 4(m - 1) = 0$,即$(m - 2)^2 = 0$,解得$m_1 = m_2 = 2$. 当$m = 2$时,方程为$x^2 - 2x + 1 = 0$,解得$x_1 = x_2 = 1$,$\therefore$ 此菱形的边长为 1.
5 思想方法分类讨论 [中]阅读材料,解答问题:
解方程:$x^2 - |x - 1| - 1 = 0$.
解:①当$x - 1 ≥ 0$,即$x ≥ 1$时,$x^2 - (x - 1) - 1 = 0$,
$x^2 - x = 0$,
解得$x_1 = 0$(不合题意,舍去),$x_2 = 1$.
②当$x - 1 ＜ 0$,即$x ＜ 1$时,$x^2 + (x - 1) - 1 = 0$,$x^2 + x - 2 = 0$,
解得$x_3 = 1$(不合题意,舍去),$x_4 = -2$.
综上所述,原方程的解是$x = 1$或$x = -2$.
依照上述解法,解方程:$x^2 + 2|x + 2| - 4 = 0$.
解方程:$x^2 - |x - 1| - 1 = 0$.
解:①当$x - 1 ≥ 0$,即$x ≥ 1$时,$x^2 - (x - 1) - 1 = 0$,
$x^2 - x = 0$,
解得$x_1 = 0$(不合题意,舍去),$x_2 = 1$.
②当$x - 1 ＜ 0$,即$x ＜ 1$时,$x^2 + (x - 1) - 1 = 0$,$x^2 + x - 2 = 0$,
解得$x_3 = 1$(不合题意,舍去),$x_4 = -2$.
综上所述,原方程的解是$x = 1$或$x = -2$.
依照上述解法,解方程:$x^2 + 2|x + 2| - 4 = 0$.
答案
①当$x + 2 ≥ 0$,即$x ≥ -2$时,$x^2 + 2(x + 2) - 4 = 0$,$x^2 + 2x = 0$,解得$x_1 = 0$,$x_2 = -2$; ②当$x + 2 ＜ 0$,即$x ＜ -2$时,$x^2 - 2(x + 2) - 4 = 0$,$x^2 - 2x - 8 = 0$,解得$x_3 = 4$(不合题意,舍去),$x_4 = -2$(不合题意,舍去). 综上所述,原方程的解是$x = 0$或$x = -2$.
