《2025年初中必刷题八年级数学上册北师大版》

1 [2025 安徽宿州调研,较难] 阅读探索：
材料一：解方程组 $\left\{\begin{array}{l} (a - 1) + 2(b + 2) = 6,\\ 2(a - 1) + (b + 2) = 6\end{array} \right.$ 时,采用了“换元法”,解法如下：
解：设 $a - 1 = x$,$b + 2 = y$,所以原方程组可化为 $\left\{\begin{array}{l} x + 2y = 6,\\ 2x + y = 6,\end{array} \right.$
解得 $\left\{\begin{array}{l} x = 2,\\ y = 2,\end{array} \right.$ 即 $\left\{\begin{array}{l} a - 1 = 2,\\ b + 2 = 2,\end{array} \right.$ 解得 $\left\{\begin{array}{l} a = 3,\\ b = 0.\end{array} \right.$
材料二：解方程组 $\left\{\begin{array}{l} 4x + 10y = 6,\enclose{circle}{1}\\ 8x + 22y = 10\enclose{circle}{2}\end{array} \right.$ 时,采用了“整体代换”的方法,解法如下：
解：将方程 $\enclose{circle}{2}$ 变形为 $2(4x + 10y) + 2y = 10$,$\enclose{circle}{3}$
把方程 $\enclose{circle}{1}$ 代入 $\enclose{circle}{3}$,得 $2×6 + 2y = 10$,解得 $y = -1$。
把 $y = -1$ 代入 $\enclose{circle}{1}$,得 $x = 4$,所以方程组的解为 $\left\{\begin{array}{l} x = 4,\\ y = -1.\end{array} \right.$
根据上述材料,解决下列问题：
(1) 换元法 ① 运用“换元法”解方程组 $\left\{\begin{array}{l} (\frac{a}{4} - 1) + 2(\frac{b}{3} + 2) = 4,\\ 2(\frac{a}{4} - 1) + (\frac{b}{3} + 2) = 5.\end{array} \right.$
② 若关于 $x$,$y$ 的方程组 $\left\{\begin{array}{l} a_1x + b_1y = c_1,\\ a_2x + b_2y = c_2\end{array} \right.$ 的解为 $\left\{\begin{array}{l} x = 10,\\ y = 6,\end{array} \right.$ 用“换元法”求关于 $m$,$n$ 的方程组 $\left\{\begin{array}{l} 5a_1(m - 3) + 3b_1(n + 2) = c_1,\\ 5a_2(m - 3) + 3b_2(n + 2) = c_2\end{array} \right.$ 的解。
(2) 整体代换 已知实数 $x$,$y$,$z$,满足 $\left\{\begin{array}{l} 3x - 2z + 12y = 47,\enclose{circle}{1}\\ 2x + z + 8y = 36,\enclose{circle}{2}\end{array} \right.$ 用“整体代换”的方法求 $z$ 的值。

答案： 【解】
(1)①运用“换元法”解方程组$\left\{\begin{array}{l} (\frac{a}{4} - 1) + 2(\frac{b}{3} + 2) = 4,\\ 2(\frac{a}{4} - 1) + (\frac{b}{3} + 2) = 5.\end{array}\right. $设$\frac {a}{4}-1=x,\frac {b}{3}+2=y$,所以原方程组可以化为$\left\{\begin{array}{l} x+2y=4,①\\ 2x+y=5,②\end{array}\right. $②-①×2,得$-3y=-3$,解得$y=1$.把$y=1$代入①,得$x+2=4$,解得$x=2$,所以方程组的解为$\left\{\begin{array}{l} x=2,\\ y=1,\end{array}\right. $即$\left\{\begin{array}{l} \frac {a}{4}-1=2,\\ \frac {b}{3}+2=1,\end{array}\right. $解得$\left\{\begin{array}{l} a=12,\\ b=-3,\end{array}\right. $所以原方程组的解为$\left\{\begin{array}{l} a=12,\\ b=-3.\end{array}\right.$②若关于$x$,$y$的方程组$\left\{\begin{array}{l} a_1x + b_1y = c_1,\\ a_2x + b_2y = c_2\end{array} \right.$的解为$\left\{\begin{array}{l} x = 10,\\ y = 6,\end{array} \right.$用“换元法”求关于$m$,$n$的方程组$\left\{\begin{array}{l} 5a_1(m - 3) + 3b_1(n + 2) = c_1,\\ 5a_2(m - 3) + 3b_2(n + 2) = c_2\end{array} \right.$的解。设$\left\{\begin{array}{l} 5(m-3)=x,\\ 3(n+2)=y,\end{array}\right. $则关于m,n的方程组可化为$\left\{\begin{array}{l} a_{1}x+b_{1}y=c_{1},\\ a_{2}x+b_{2}y=c_{2}.\end{array}\right. $因为关于x,y的方程组的解为$\left\{\begin{array}{l} x=10,\\ y=6,\end{array}\right. $所以$\left\{\begin{array}{l} 5(m-3)=10,\\ 3(n+2)=6,\end{array}\right. $解得$\left\{\begin{array}{l} m=5,\\ n=0.\end{array}\right.$
(2)整体代换 已知实数$x$,$y$,$z$,满足$\left\{\begin{array}{l} 3x - 2z + 12y = 47,\enclose{circle}{1}\\ 2x + z +8y = 36,\enclose{circle}{2}\end{array} \right.$用“整体代换”的方法求$z$的值。将方程①变形为$\frac {3}{2}(2x+z+8y)-\frac {7}{2}z=47$,③ 将方程②代入③,得$\frac {3}{2}×36-\frac {7}{2}z=47$,解得$z=2.$

2 [2025 四川巴中质检,较难] 阅读下列材料,并解答问题。
解方程组 $\left\{\begin{array}{l} 14x + 15y = 16,\enclose{circle}{1}\\ 17x + 18y = 19\enclose{circle}{2}\end{array} \right.$ 时,由于 $x$,$y$ 的系数及常数项的值较大,如果用常规的代入消元法或加减消元法求解,那么计算量较大,且易出现运算错误,而采用下面的解法则比较简单：
$\enclose{circle}{2} - \enclose{circle}{1}$,得 $3x + 3y = 3$,所以 $x + y = 1$,$\enclose{circle}{3}$
$\enclose{circle}{3}×14$,得 $14x + 14y = 14$,$\enclose{circle}{4}$
$\enclose{circle}{1} - \enclose{circle}{4}$,得 $y = 2$,把 $y = 2$ 代入 $\enclose{circle}{3}$,得 $x = -1$。
所以原方程组的解为 $\left\{\begin{array}{l} x = -1,\\ y = 2.\end{array} \right.$
(1) 叠减法 请你采用上述方法解方程组 $\left\{\begin{array}{l} 2025x + 2024y = 1,\\ 2023x + 2022y = 1.\end{array} \right.$
(2) 叠加法 我们把形如 $\left\{\begin{array}{l} mx + ny = a,\\ nx + my = b\end{array} \right.$(未知数系数交换)的方程组称为轮换式方程组。请解轮换式方程组 $\left\{\begin{array}{l} 2025x + 2024y = 2026,\\ 2024x + 2025y = 2023.\end{array} \right.$

答案： 【解】
(1)叠减法 解方程组$\left\{\begin{array}{l} 2025x + 2024y = 1,\\ 2023x + 2022y = 1.\end{array} \right. $$\left\{\begin{array}{l} 2025x+2024y=1,①\\ 2023x+2022y=1,②\end{array}\right. $①-②,得$2x+2y=0$,即$x+y=0$.③ ③×2022,得$2022x+2022y=0$.④ ②-④,得$x=1$,把$x=1$代入③,得$y=-1$,所以原方程组的解为$\left\{\begin{array}{l} x=1,\\ y=-1.\end{array}\right.$
(2)叠加法 解轮换式方程组$\left\{\begin{array}{l} 2025x + 2024y = 2026,\\ 2024x + 2025y = 2023.\end{array} \right. $$\left\{\begin{array}{l} 2025x+2024y=2026,①\\ 2024x+2025y=2023,②\end{array}\right. $①+②,得$4049x+4049y=4049$,即$x+y=1$.③ ③×2024,得$2024x+2024y=2024$.④ ②-④,得$y=-1$,将$y=-1$代入③,得$x=2$,所以原方程组的解为$\left\{\begin{array}{l} x=2,\\ y=-1.\end{array}\right. $