《2025年通城学典暑期升级训练七年级数学》

1. 若$(k - 2)x + 1 = 0$是关于$x$的一元一次方程，则$k$的值不可能是 (C)
A. $-1$
B. $0$
C. $2$
D. $-2$

答案： C

2. 若$x^{m - 2} - y^{n + 4} = 21$是关于$x$，$y$的二元一次方程，则$m + n$值为______0。

答案： 0

3. 已知$3x^{2m+1}-4y^{3m+2n}= 1$是关于$x$,$y$的二元一次方程,则$2m-3n= $$-\frac{3}{2}$.

答案： $-\frac{3}{2}$

4. (眉山中考)已知关于$x$,$y$的二元一次方程组$\left\{\begin{array}{l} 3x-y= 4m+1,\\ x+y= 2m-5\end{array} \right.$的解满足$x-y= 4$,则$m$的值为 (B)
A. $0$
B. $1$
C. $2$
D. $3$

答案： B

5. 若$x= 1是关于x的方程-2mx+n-1= 0$的解,则$2024+n-2m$的值为 (C)
A. $2023$
B. $2024$
C. $2025$
D. $2026$

答案： C

6. 如果关于$x$,$y$的二元一次方程组$\left\{\begin{array}{l} x+y= 3,\\ 2x-ay= 5\end{array} \right.$的解是$\left\{unitable2 \right.$ 那么$a^{b}$的值为______1____.

答案： 1

7. 已知$\left\{\begin{array}{l} x= 2,\\ y= 1\end{array} \right. $是关于x,y的二元一次方程组$\left\{ \begin{array} { l } { 2 x + ( m - 1 ) y = 2 , } \\ { n x + y = 1 } \end{array} \right. $的解,求$(m+n)^{2024}$的值.
把$\left\{\begin{array}{l}x = 2,\\ y = 1\end{array}\right.$代入方程组，得$\left\{\begin{array}{l}4 + m - 1 = 2,\\ 2n + 1 = 1,\end{array}\right.$解得$\left\{\begin{array}{l}m = - 1,\\ n = 0.\end{array}\right.$
$\therefore (m + n)^{2024}$的值为1。

答案： 把$\left\{\begin{array}{l}x = 2,\\ y = 1\end{array}\right.$代入方程组，得$\left\{\begin{array}{l}4 + m - 1 = 2,\\ 2n + 1 = 1,\end{array}\right.$解得$\left\{\begin{array}{l}m = - 1,\\ n = 0.\end{array}\right.$
$\therefore (m + n)^{2024} = 1$。

8. 新考法 新定义题 如果关于x,y的二元一次方程组$\left\{\begin{array}{l} a_{1}x+b_{1}y= c_{1},\\ a_{2}x+b_{2}y= c_{2}\end{array} \right. $与$\left\{\begin{array}{l} a_{3}x+b_{3}y= c_{3},\\ a_{4}x+b_{4}y= c_{4}\end{array} \right. $ 的解分别是$\begin{cases}{x=y_0,}\\{y=x_0}\end{cases}$与那么称这两个二元一次方程组的解是对称解.已知关于x,y的方程组$\left\{ \begin{array} { l } { 2 x + 5 y = - 6 }, \\ { a x - b y = - 4 } \end{array} \right. \text { 与 } \left\{ \begin{array} { l } { x - 4 y = 23 }, \\ { b x + a y = 8 } \end{array} \right.$的解是对称解。
$\because$关于$x$，$y$的方程组$\left\{\begin{array}{l}2x + 5y = - 6,\\ ax - by = - 4\end{array}\right.$与方程组$\left\{\begin{array}{l}x - 4y = 23,\\ bx + ay = 8\end{array}\right.$的解是对称解，$\therefore$易得方程组$\left\{\begin{array}{l}2x + 5y = - 6,\\ y - 4x = 23,\end{array}\right.$解得$\left\{\begin{array}{l}x = {-\frac{11}{2}},\\ y = {1}.\end{array}\right.$ $\therefore$第一个方程组的解是$\left\{\begin{array}{l}x = {-\frac{11}{2}},\\ y = {1},\end{array}\right.$第二个方程组的解是$\left\{\begin{array}{l}x = {1},\\ y = {-\frac{11}{2}}.\end{array}\right.$把$\left\{\begin{array}{l}x = {-\frac{11}{2}},\\ y = {1}\end{array}\right.$代入$ax - by = - 4$，得$-\frac{11}{2}a - b = - 4$①。把$\left\{\begin{array}{l}x = {1},\\ y = {-\frac{11}{2}}\end{array}\right.$代入$bx + ay = 8$，得$b - \frac{11}{2}a = 8$②。由①②，得$\left\{\begin{array}{l}a = {-\frac{4}{11}},\\ b = {6}.\end{array}\right.$ $\therefore a = {-\frac{4}{11}}$，$b = {6}$，第一个方程组的解为$\left\{\begin{array}{l}x = {-\frac{11}{2}},\\ y = {1},\end{array}\right.$第二个方程组的解为$\left\{\begin{array}{l}x = {1},\\ y = {-\frac{11}{2}}.\end{array}\right.$

答案： $\because$关于$x$，$y$的方程组$\left\{\begin{array}{l}2x + 5y = - 6,\\ ax - by = - 4\end{array}\right.$与方程组$\left\{\begin{array}{l}x - 4y = 23,\\ bx + ay = 8\end{array}\right.$的解是对称解，$\therefore$易得方程组$\left\{\begin{array}{l}2x + 5y = - 6,\\ y - 4x = 23,\end{array}\right.$解得$\left\{\begin{array}{l}x = -\frac{11}{2},\\ y = 1.\end{array}\right.$ $\therefore$第一个方程组的解是$\left\{\begin{array}{l}x = -\frac{11}{2},\\ y = 1,\end{array}\right.$第二个方程组的解是$\left\{\begin{array}{l}x = 1,\\ y = -\frac{11}{2}.\end{array}\right.$把$\left\{\begin{array}{l}x = -\frac{11}{2},\\ y = 1\end{array}\right.$代入$ax - by = - 4$，得$-\frac{11}{2}a - b = - 4$①。把$\left\{\begin{array}{l}x = 1,\\ y = -\frac{11}{2}\end{array}\right.$代入$bx + ay = 8$，得$b - \frac{11}{2}a = 8$②。由①②，得$\left\{\begin{array}{l}a = -\frac{4}{11},\\ b = 6.\end{array}\right.$ $\therefore a = -\frac{4}{11}$，$b = 6$，第一个方程组的解为$\left\{\begin{array}{l}x = -\frac{11}{2},\\ y = 1,\end{array}\right.$第二个方程组的解为$\left\{\begin{array}{l}x = 1,\\ y = -\frac{11}{2}.\end{array}\right.$