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1 (2025 南京玄武月考)与方程组$\begin{cases}x + 2y - 3 = 0,\\2x + y = 0\end{cases}$有完全相同解的方程是( )
A.$x + 2y = 3$
B.$2x + y = 0$
C.$(x + 2y - 3)(2x + y) = 0$
D.$x + 2y - 3 + (2x + y)^2 = 0$
A.$x + 2y = 3$
B.$2x + y = 0$
C.$(x + 2y - 3)(2x + y) = 0$
D.$x + 2y - 3 + (2x + y)^2 = 0$
答案
1. D
2 (2025 南通崇川期中)已知关于$x$，$y$的方程组$\begin{cases}2x - 3y = 3,\\mx + ny = -1\end{cases}$和$\begin{cases}2mx + 3ny = 3,\\3x + 2y = 11\end{cases}$的解相同，求$(3m + n)^{2025}$的值。
答案
2. 解:根据题意,得$\{\begin{array}{l} 2x - 3y = 3①,\\ 3x + 2y = 11②,\end{array} $
由①×2 + ②×3,得$13x = 39$,解得$x = 3$,
将$x = 3$代入①,得$y = 1$,所以$\{\begin{array}{l} x = 3,\\ y = 1,\end{array} $
将$\{\begin{array}{l} x = 3,\\ y = 1\end{array} $代入$\{\begin{array}{l} 2mx + 3ny = 3,\\ mx + ny = - 1,\end{array} $得$\{\begin{array}{l} 2m + n = 1,\\ 3m + n = - 1,\end{array} $
解得$\{\begin{array}{l} m = - 2,\\ n = 5,\end{array} $
所以$(3m + n)^{2025} = [3×(- 2) + 5]^{2025} = (- 1)^{2025} = - 1$.
由①×2 + ②×3,得$13x = 39$,解得$x = 3$,
将$x = 3$代入①,得$y = 1$,所以$\{\begin{array}{l} x = 3,\\ y = 1,\end{array} $
将$\{\begin{array}{l} x = 3,\\ y = 1\end{array} $代入$\{\begin{array}{l} 2mx + 3ny = 3,\\ mx + ny = - 1,\end{array} $得$\{\begin{array}{l} 2m + n = 1,\\ 3m + n = - 1,\end{array} $
解得$\{\begin{array}{l} m = - 2,\\ n = 5,\end{array} $
所以$(3m + n)^{2025} = [3×(- 2) + 5]^{2025} = (- 1)^{2025} = - 1$.
3 (2025 宿近期末)解方程组$\begin{cases}ax + by = 2,\\cx - 7y = 8\end{cases}$时，一名学生将$c$看错而得到$\begin{cases}x = -2,\\y = 2,\end{cases}$而正确的解是$\begin{cases}x = 3,\\y = -2,\end{cases}$则$a$，$b$，$c$的值为( )
A.不能确定
B.$a = 4$，$b = 5$，$c = -2$
C.$a$，$b$不能确定，$c = -2$
D.$a = 4$，$b = 7$，$c = 2$
A.不能确定
B.$a = 4$，$b = 5$，$c = -2$
C.$a$，$b$不能确定，$c = -2$
D.$a = 4$，$b = 7$，$c = 2$
答案
3. B
4 在解方程组$\begin{cases}ax + 5y = 15,\\4x - by = -2\end{cases}$时，由于粗心，甲看错了方程组中的$a$，得到的解为$\begin{cases}x = -3,\\y = -1,\end{cases}$乙看错了方程组中的$b$，得到的解为$\begin{cases}x = 5,\\y = 4.\end{cases}$
(1) 甲将$a$看成了什么，乙将$b$看成了什么；
(2) 求出原方程组的正确解。
(1) 甲将$a$看成了什么，乙将$b$看成了什么；
(2) 求出原方程组的正确解。
答案
4. 解:(1)将$\{\begin{array}{l} x = - 3,\\ y = - 1\end{array} $代入原方程组,得
$\{\begin{array}{l} - 3a - 5 = 15,\\ 4×(- 3) + b = - 2,\end{array} $解得$\{\begin{array}{l} a_{\mathrm{错}} = - \dfrac{20}{3},\\ b = 10.\end{array} $
将$\{\begin{array}{l} x = 5,\\ y = 4\end{array} $代入原方程组,得$\{\begin{array}{l} 5a + 20 = 15,\\ 20 - 4b = - 2,\end{array} $
解得$\{\begin{array}{l} a = - 1,\\ b_{\mathrm{错}} = \dfrac{11}{2},\end{array} $
所以甲将$a$看成了$- \dfrac{20}{3}$,乙将$b$看成了$\dfrac{11}{2}$.
(2)由(1)可知原方程组中$a = - 1$,$b = 10$,
所以原方程组为$\{\begin{array}{l} - x + 5y = 15,\\ 4x - 10y = - 2,\end{array} $解得$\{\begin{array}{l} x = 14,\\ y = \dfrac{29}{5}.\end{array} $
$\{\begin{array}{l} - 3a - 5 = 15,\\ 4×(- 3) + b = - 2,\end{array} $解得$\{\begin{array}{l} a_{\mathrm{错}} = - \dfrac{20}{3},\\ b = 10.\end{array} $
将$\{\begin{array}{l} x = 5,\\ y = 4\end{array} $代入原方程组,得$\{\begin{array}{l} 5a + 20 = 15,\\ 20 - 4b = - 2,\end{array} $
解得$\{\begin{array}{l} a = - 1,\\ b_{\mathrm{错}} = \dfrac{11}{2},\end{array} $
所以甲将$a$看成了$- \dfrac{20}{3}$,乙将$b$看成了$\dfrac{11}{2}$.
(2)由(1)可知原方程组中$a = - 1$,$b = 10$,
所以原方程组为$\{\begin{array}{l} - x + 5y = 15,\\ 4x - 10y = - 2,\end{array} $解得$\{\begin{array}{l} x = 14,\\ y = \dfrac{29}{5}.\end{array} $
5 (2025 盐城东台月考)若关于$x$，$y$的方程组$\begin{cases}2x - y = 5k + 6,\\4x + 7y = k\end{cases}$的解满足$x + y = 2024$，则$k$的值为( )
A.$2022$
B.$2023$
C.$2024$
D.$2025$
A.$2022$
B.$2023$
C.$2024$
D.$2025$
答案
5. B
