《2025年名校课堂七年级数学上册湘教版广西专版》

1.【阅读材料】(2023·南宁四十七中期末)阅读材料:善于思考的乐乐同学在解方程组$\left\{\begin{array}{l} 3(m+5)-2(n+3)= -1,\\ 3(m+5)+2(n+3)= 7\end{array} \right. $时,采用了一种“整体换元”的解法.把$m+5,n+3$看成一个整体,设$x= m+5,y= n+3$,则原方程组可化为$\left\{\begin{array}{l} 3x-2y= -1,\\ 3x+2y= 7,\end{array} \right. 解得\left\{\begin{array}{l} x= 1,\\ y= 2.\end{array} \right. 所以\left\{\begin{array}{l} m+5= 1,\\ n+3= 2,\end{array} \right. 解得\left\{\begin{array}{l} m= -4,\\ n= -1.\end{array} \right. $
(1)【学以致用】模仿乐乐同学的“整体换元”法,解方程组:$\left\{\begin{array}{l} \frac {x+y}{3}+\frac {x-y}{5}= 4,\\ \frac {x+y}{3}-\frac {x-y}{5}= -2;\end{array} \right. $
(2)【拓展提升】已知关于$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. 的解为\left\{\begin{array}{l} x= 3,\\ y= 4,\end{array} \right. 请直接写出关于m,n的二元一次方程组\left\{\begin{array}{l} a_{1}(m+2)-b_{1}n= c_{1},\\ a_{2}(m+2)-b_{2}n= c_{2}\end{array} \right. $的解:$\left\{\begin{array}{l} m=1,\\ n=4\end{array}\right.$.
(1)令$m=\frac{x+y}{3},n=\frac{x-y}{5}$,则原方程组可化为$\left\{\begin{array}{l} m+n=4,\\ m-n=-2,\end{array}\right.$解得$\left\{\begin{array}{l} m=1,\\ n=3.\end{array}\right.$所以$\left\{\begin{array}{l} \frac{x+y}{3}=1,\\ \frac{x-y}{5}=3,\end{array}\right.$解得$\left\{\begin{array}{l} x=9,\\ y=-6.\end{array}\right.$

答案：
(1)令$m=\frac{x+y}{3},n=\frac{x-y}{5}$,则原方程组可化为$\left\{\begin{array}{l} m+n=4,\\ m-n=-2,\end{array}\right.$解得$\left\{\begin{array}{l} m=1,\\ n=3.\end{array}\right.$所以$\left\{\begin{array}{l} \frac{x+y}{3}=1,\\ \frac{x-y}{5}=3,\end{array}\right.$解得$\left\{\begin{array}{l} x=9,\\ y=-6.\end{array}\right.$
(2)$\left\{\begin{array}{l} m=1,\\ n=4\end{array}\right.$

2. 当$x= 1$时,整式$ax^{3}+bx-2$的值为2024,则当$x= -1$时,整式$ax^{3}+bx-2$的值是-2028.

答案： -2028

3. 新考向 阅读理解(2023·南宁良庆区期末)阅读理解:
在数学课上,李老师遇到下面问题:已知$x,y满足方程组\left\{\begin{array}{l} x+3y= -1,\\ 3x+y= 5,\end{array} \right. 求x+y$的值.
小红:把方程组解出来,再求$x+y$的值.
小刚:把两个方程直接相加,得$4x+4y= 4$,方程两边同时除以$4$,得$x+y= 1$.
李老师对两位同学的讲解进行点评:小刚同学的思路体现了数学中“整体思想”的运用.
请参考小红或小刚同学的做法,解决下面的问题.
(1)已知关于$x,y的方程组\left\{\begin{array}{l} 2x+y= 2a+1,\\ x+2y= 5-5a\end{array} \right. 的解满足x+y= -3$,求$a$的值;
(2)运用“整体思想”解答:
若方程组$\left\{\begin{array}{l} ax+y= b,\\ x-by= a\end{array} \right. 的解是\left\{\begin{array}{l} x= 1,\\ y= 1,\end{array} \right. 求(a+b)^{2}-(a-b)(a+b)$的值.

答案：
(1)$\left\{\begin{array}{l} 2x+y=2a+1,\enclose{circle}{1}\\ x+2y=5-5a,\enclose{circle}{2}\end{array}\right. (\enclose{circle}{1}+\enclose{circle}{2})÷3$,得$x+y=2-a$.又因为$x+y=-3$,所以$2-a=-3$.解得$a=5$.
(2)将$\left\{\begin{array}{l} x=1,\\ y=1\end{array}\right.$代入原方程组,得$\left\{\begin{array}{l} a+1=b,\\ 1-b=a,\end{array}\right.$整理,得$\left\{\begin{array}{l} a-b=-1,\\ a+b=1,\end{array}\right.$所以$(a+b)^{2}-(a-b)(a+b)=1^{2}-(-1)×1=1+1=2$

4. 一个四位数,它的个位数字是$8$,若把这个数字调到千位上,其他数字向后顺移,得到新的四位数比原来的四位数大$117$,求原来的四位数.

答案： 解:设原来的四位数去掉个位数字后的三位数为$x$,于是原来的四位数为$10x+8$,新的四位数可表示为$8000+x$,由题意,得$8000+x=10x+8+117$,解得$x=875$.答:原来的四位数是8758