《2025年通城学典暑期升级训练七年级数学》

9. 新考法 阅读理解 甲、乙两人同时解方程组$\left\{\begin{array}{l} mx+y= 5①,\\ 2x-ny= 13②,\end{array} \right. $甲看错了①中的$m$,解得$\begin{cases}{x=\frac 72,}\\{y=-2}\end{cases}$乙看错了②中的$n$,解得$\begin{cases}{x=3,}\\{y=-7}\end{cases} $。把$\left\{\begin{array}{l}x = \frac{7}{2},\\ y = - 2\end{array}\right.$代入②，得$7 + 2n = 13$，解得$n = 3$。把$\left\{\begin{array}{l}x = 3,\\ y = - 7\end{array}\right.$代入①，得$3m - 7 = 5$，解得$m = 4$。$\therefore \sqrt{m^{2} + n^{2}} =$$5$。

答案： 把$\left\{\begin{array}{l}x = \frac{7}{2},\\ y = - 2\end{array}\right.$代入②，得$7 + 2n = 13$，解得$n = 3$。把$\left\{\begin{array}{l}x = 3,\\ y = - 7\end{array}\right.$代入①，得$3m - 7 = 5$，解得$m = 4$。
$\therefore \sqrt{m^{2} + n^{2}} = 5$。

10. 若关于$x$,$y$的二元一次方程组$\left\{\begin{array}{l} ax + 2y = 1\\ 3x + y = 3\end{array}\right.$有唯一解,则下列结论正确的是 (B)
A. $a\neq 0$
B. $a\neq 6$
C. $a = 0$
D. $a$为任意数

答案： B

11. 已知$m$为整数,关于$x$,$y$的二元一次方程组$\left\{\begin{array}{l} 4x - 3y = 6,\\ 6x + my = 26\end{array} \right. $有正整数解,则$m$的值为4 或 - 4.

答案： 4 或 - 4

12. 若关于x,y的方程组$\left\{\begin{array}{l} ax+by= 4,\\ 3x-y= 2\end{array} \right. $与$\left\{ \begin{array} { l } { x + 2 y = 1 , } \\ { a x - b y = - 2 } \end{array} \right.$有相同的解,求$a$,$b$的值.
$\because$关于$x$，$y$的方程组$\left\{\begin{array}{l}ax + by = 4,\\ 3x - y = 2\end{array}\right.$与$\left\{\begin{array}{l}x + 2y = 1,\\ ax - by = - 2\end{array}\right.$有相同的解，$\therefore$方程组$\left\{\begin{array}{l}3x - y = 2,\\ x + 2y = 1\end{array}\right.$的解也是它们的解，解得$\left\{\begin{array}{l}x = \frac{5}{7},\\ y = \frac{1}{7}.\end{array}\right.$将解代入其他两个方程，得$\left\{\begin{array}{l}\frac{5}{7}a + \frac{1}{7}b = 4,\\ \frac{5}{7}a - \frac{1}{7}b = - 2,\end{array}\right.$解得$\left\{\begin{array}{l}a = \frac{7}{5},\\ b = 21.\end{array}\right.$

答案： $\because$关于$x$，$y$的方程组$\left\{\begin{array}{l}ax + by = 4,\\ 3x - y = 2\end{array}\right.$与$\left\{\begin{array}{l}x + 2y = 1,\\ ax - by = - 2\end{array}\right.$有相同的解，$\therefore$方程组$\left\{\begin{array}{l}3x - y = 2,\\ x + 2y = 1\end{array}\right.$的解也是它们的解，解得$\left\{\begin{array}{l}x = \frac{5}{7},\\ y = \frac{1}{7}.\end{array}\right.$将解代入其他两个方程，得$\left\{\begin{array}{l}\frac{5}{7}a + \frac{1}{7}b = 4,\\ \frac{5}{7}a - \frac{1}{7}b = - 2,\end{array}\right.$解得$\left\{\begin{array}{l}a = \frac{7}{5},\\ b = 21.\end{array}\right.$

13. 已知$4x^{2k+3}+11+k^{2}＞4$是关于$x$的一元一次不等式,则$k$的值为 (C)
A.$0$
B.$1$
C.$-1$
D.$-2$

答案： C

14. 已知关于$x$的不等式$\frac {2m - mx}{2}＞\frac {1}{2}x - 1$是一元一次不等式，则$m$的取值范围是__$m\neq - 1$__.

答案： $m\neq - 1$

15. 若关于$x$的一元一次不等式$\frac {m - 2x}{3} \leq -2$的解集为$x \geq 5$，则$m$的值为 (D)
A.$-2$
B.$2$
C.$-4$
D.$4$

答案： D

16. 已知关于$x$的一元一次不等式组$\left\{\begin{array}{l} 2+2(1-x)＞13-5x,\\ x-k＞-2\end{array} \right.$的解集是$x＞3$，则$k$的取值范围是 (B)
A.$k＜5$
B.$k\leq 5$
C.$k＞5$
D.$k\geq 5$

答案： B