《2025年暑假乐园现代教育出版社七年级数学人教版》

1. 已知$\left\{ \begin{array} { l } { x = 2 m, } \\ { y = n - 1 } \end{array} \right.$是二元一次方程组$\left\{ \begin{array} { l } { 2 x - 3 y = 2, } \\ { 4 x + 5 y = 26 } \end{array} \right.$$\begin{array} { l } { ① } \\ { ② } \end{array}$的解，求式子$m ^ { n } - 1$的值.
解：② - ①×2，$11y = 22$. 解得 $y = $2.
把 $y = $2代入①，得 $2x - 3×$2$ = 2$. 解得 $x = $4.
原方程组的解为 $\left\{\begin{array}{l} x = $4,\\ y = $2.\end{array}\right.$
所以 $2m = $4，$n - 1 = $2，即 $m = $2，$n = $3.
故 $m^n - 1 = 2^3 - 1 = 8 - 1 = $7.

答案： 解：② - ①×2，$11y = 22$. 解得 $y = 2$.
把 $y = 2$ 代入①，得 $2x - 3×2 = 2$. 解得 $x = 4$.
原方程组的解为 $\left\{\begin{array}{l} x = 4,\\ y = 2.\end{array}\right.$
所以 $2m = 4$，$n - 1 = 2$，即 $m = 2$，$n = 3$.
故 $m^n - 1 = 2^3 - 1 = 8 - 1 = 7$.

2. 小明、小亮两人在解同一个二元一次方程组$\left\{ \begin{array} { l } { 2 x + \otimes y = 6, } \\ { x + \oplus y = 14, } \end{array} \right.$$\begin{array} { l } { ① } \\ { ② } \end{array}$时，由于小明看错了方程①中$y$的系数，得到二元一次方程组的解为$\left\{ \begin{array} { l } { x = 10, } \\ { y = \frac { 4 } { 3 } ; } \end{array} \right.$小亮看错了方程②中$y$的系数，得到二元一次方程组的解为$\left\{ \begin{array} { l } { x = 2, } \\ { y = 2. } \end{array} \right.$试求出原二元一次方程组并求解.
解：把 $\left\{\begin{array}{l} x = 10,\\ y = \frac{4}{3}\end{array}\right.$ 代入方程②，得 $10 + \oplus × \frac{4}{3} = 14$. 解得 $\oplus =$3.
把 $\left\{\begin{array}{l} x = 2,\\ y = 2\end{array}\right.$ 代入方程①，得 $2×2 + \otimes ×2 = 6$. 解得 $\otimes =$1.
所以原方程组为 $\left\{\begin{array}{l} 2x + y = 6, ①\\ x + 3y = 14. ②\end{array}\right.$
由①，得 $y = 6 - 2x$. ③
把③代入②，得 $x + 3(6 - 2x) = 14$.
解得 $x =$$\frac{4}{5}$.
把 $x = \frac{4}{5}$ 代入③，得 $y = 6 - 2×\frac{4}{5} =$$\frac{22}{5}$.
所以原方程组的解是 $\left\{\begin{array}{l} x =$$\frac{4}{5}$,\\ y =$$\frac{22}{5}$.\end{array}\right.$

答案： 解：把 $\left\{\begin{array}{l} x = 10,\\ y = \frac{4}{3}\end{array}\right.$ 代入方程②，得 $10 + \oplus × \frac{4}{3} = 14$. 解得 $\oplus = 3$.
把 $\left\{\begin{array}{l} x = 2,\\ y = 2\end{array}\right.$ 代入方程①，得 $2×2 + \otimes ×2 = 6$. 解得 $\otimes = 1$.
所以原方程组为 $\left\{\begin{array}{l} 2x + y = 6, ①\\ x + 3y = 14. ②\end{array}\right.$
由①，得 $y = 6 - 2x$. ③
把③代入②，得 $x + 3(6 - 2x) = 14$.
解得 $x = \frac{4}{5}$.
把 $x = \frac{4}{5}$ 代入③，得 $y = 6 - 2×\frac{4}{5} = \frac{22}{5}$.
所以原方程组的解是 $\left\{\begin{array}{l} x = \frac{4}{5},\\ y = \frac{22}{5}.\end{array}\right.$