《2025年天府前沿八年级数学上册北师大版》

9. (1)(2025·编写)若关于x,y的二元一次方程组$\left\{\begin{array}{l} 3x+2y= m+1,\\ 2x+3y= m\end{array} \right. 的解满足x+y= 3$,求m的值.
(2)(2024·怀远)若单项式$2a^{m}b^{2}与-a^{4}b^{n}$是同类项,求方程组$\left\{\begin{array}{l} x= my,\\ nx+y= mn\end{array} \right. $的解.

答案：
(1)【解】$\left\{ \begin{array} { l } { 3 x + 2 y = m + 1, \textcircled { 1 } } \\ { 2 x + 3 y = m. \textcircled { 2 } } \end{array} \right.$
由①+②,得$5 x + 5 y = 2 m + 1$,即$x + y = \frac { 2 m + 1 } { 5 }$.
$\because x + y = 3, \therefore \frac { 2 m + 1 } { 5 } = 3$,解得$m = 7$.
(2)【解】若单项式$2 a ^ { m } b ^ { 2 }$与$- a ^ { 4 } b ^ { n }$是同类项,
则$m = 4, n = 2$,
$\therefore$方程组$\left\{ \begin{array} { l } { x = m y, } \\ { n x + y = m n } \end{array} \right.$为$\left\{ \begin{array} { l } { x = 4 y, \textcircled { 1 } } \\ { 2 x + y = 8, \textcircled { 2 } } \end{array} \right.$
把①代入②,得$2 × 4 y + y = 8$,解得$y = \frac { 8 } { 9 }$.
把$y = \frac { 8 } { 9 }$代入①,得$x = \frac { 32 } { 9 }$,
$\therefore$方程组$\left\{ \begin{array} { l } { x = m y, } \\ { n x + y = m n } \end{array} \right.$的解为$\left\{ \begin{array} { l } { x = \frac { 32 } { 9 }, } \\ { y = \frac { 8 } { 9 }. } \end{array} \right.$

10. (1)(2025·泰兴)某班级为准备毕业联欢会,想购买价格分别为2元、4元和10元的三种物品,每种物品至少购买一件,共16件,恰好用50元,若2元的奖品购买x件,求符合要求的x的值.
(2)(2025·黄冈)某商场新进一种服装,每套服装售价100元,若将裤子降价10%,上衣涨价5%,调价后这套服装的单价和比原来提高了2%,这套服装原来裤子和上衣的单价分别是多少?

答案：
(1)【解】设分别购买4元和10元的物品y件,z件.
由题意,得$\left\{ \begin{array} { l } { x + y + z = 16, } \\ { 2 x + 4 y + 10 z = 50, } \end{array} \right.$解得$\left\{ \begin{array} { l } { x = 7 + 3 z, } \\ { y = 9 - 4 z. } \end{array} \right.$
当$z = 1$时,$x = 7 + 3 = 10, y = 9 - 4 = 5$;
当$z = 2$时,$x = 7 + 6 = 13, y = 9 - 8 = 1$;
当$z = 3$时,$y = 9 - 12 = - 3 ＜ 0$(不合题意).
综上所述,$x = 10$或13.
(2)【解】设裤子原来的单价是x元,上衣原来的单价是y元.
由题意,得$\left\{ \begin{array} { l } { x + y = 100, } \\ { ( 1 - 10 \% ) x + ( 1 + 5 \% ) y = 100 ( 1 + 2 \% ), } \end{array} \right.$
解得$\left\{ \begin{array} { l } { x = 20, } \\ { y = 80. } \end{array} \right.$
即这套服装原来裤子的单价是20元,上衣的单价是80元.

11. (2024·宁波)已知$x+y= 4,|x|+|y|= 7$,则$x-y$的值是$\pm 7$.

答案： $\pm 7$

12. (2025·编写)七年级(6)班有50名学生参加军训. 军训基地有6人间和4人间两种房间,若每个房间都住满,则安排这个班的学生入住的方案共有4种.

答案： 4

13. (2025·上城)已知关于x,y的方程组$\left\{\begin{array}{l} x+y= 1-a,\\ x-y= 2a+5,\end{array} \right. $则下列结论中正确的是____(填序号).
①当$a= 1$时,方程组的解也是方程$2x-y= 3.5$的解;②当$x= y$时,$a= -2.5$;③不论a取何值,$3x+y$的值始终不变;④设$z= x(y-1)$,则z的最大值为3.
②③④

答案： ②③④

14. (2024·无锡)我们把关于x,y的二元一次方程$ax+by+c= 0$的系数a,b,c称为该方程的伴随数,记作$(a,b,c)$. 例如:二元一次方程$5x-y+3= 0的伴随数是(5,-1,3)$.
(1) 二元一次方程$3x+2y= 1$的伴随数是____( 3, 2, - 1 );
(2) 已知关于x,y的二元一次方程的伴随数是$(3,m,n)$.
①若$\left\{\begin{array}{l} x= 2,\\ y= -1\end{array} \right. 和\left\{\begin{array}{l} x= -2,\\ y= 2\end{array} \right. $是该方程的两组解,求m,n的值;
②若$\left\{\begin{array}{l} x= -3,\\ y= 2\end{array} \right. $是该方程的一组解,且满足$m+n＞7$,求代数式$3m+4n$的取值范围.
(1)二元一次方程$3 x + 2 y = 1$可整理为$3 x + 2 y - 1 = 0$,$\therefore$二元一次方程$3 x + 2 y = 1$的伴随数是$( 3, 2, - 1 )$.故答案为$( 3, 2, - 1 )$.
(2)①$\because$关于x,y的二元一次方程的伴随数是$( 3, m, n )$,$\therefore$该二元一次方程为$3 x + m y + n = 0$.
将$\left\{ \begin{array} { l } { x = 2, } \\ { y = - 1 } \end{array} \right.$和$\left\{ \begin{array} { l } { x = - 2, } \\ { y = 2 } \end{array} \right.$分别代入$3 x + m y + n = 0$,得$\left\{ \begin{array} { l } { 6 - m + n = 0, } \\ { - 6 + 2 m + n = 0, } \end{array} \right.$解得$\left\{ \begin{array} { l } { m = 4, } \\ { n = - 2. } \end{array} \right.$
②将$\left\{ \begin{array} { l } { x = - 3, } \\ { y = 2 } \end{array} \right.$代入$3 x + m y + n = 0$,得$- 9 + 2 m + n = 0$,整理,得$2 m + n = 9$,
$\therefore n = 9 - 2 m, \therefore m + ( 9 - 2 m ) ＞ 7$,解得$m ＜ 2$,
$\therefore m ＜ 2$且$m \neq 0$,
$\therefore 3 m + 4 n = 3 m + 4 ( 9 - 2 m ) = 36 - 5 m ＞ 26$,
$\therefore 3 m + 4 n ＞ 26$且$3 m + 4 n \neq 36$.

答案：
(1)$( 3, 2, - 1 )$
【解】
(1)二元一次方程$3 x + 2 y = 1$可整理为$3 x + 2 y - 1 = 0$,$\therefore$二元一次方程$3 x + 2 y = 1$的伴随数是$( 3, 2, - 1 )$.故答案为$( 3, 2, - 1 )$.
(2)①$\because$关于x,y的二元一次方程的伴随数是$( 3, m, n )$,$\therefore$该二元一次方程为$3 x + m y + n = 0$.
将$\left\{ \begin{array} { l } { x = 2, } \\ { y = - 1 } \end{array} \right.$和$\left\{ \begin{array} { l } { x = - 2, } \\ { y = 2 } \end{array} \right.$分别代入$3 x + m y + n = 0$,得$\left\{ \begin{array} { l } { 6 - m + n = 0, } \\ { - 6 + 2 m + n = 0, } \end{array} \right.$解得$\left\{ \begin{array} { l } { m = 4, } \\ { n = - 2. } \end{array} \right.$
②将$\left\{ \begin{array} { l } { x = - 3, } \\ { y = 2 } \end{array} \right.$代入$3 x + m y + n = 0$,得$- 9 + 2 m + n = 0$,整理,得$2 m + n = 9$,
$\therefore n = 9 - 2 m, \therefore m + ( 9 - 2 m ) ＞ 7$,解得$m ＜ 2$,
$\therefore m ＜ 2$且$m \neq 0$,
$\therefore 3 m + 4 n = 3 m + 4 ( 9 - 2 m ) = 36 - 5 m ＞ 26$,
$\therefore 3 m + 4 n ＞ 26$且$3 m + 4 n \neq 36$.