2025年初中学霸创新题七年级数学下册华师大版


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《2025年初中学霸创新题七年级数学下册华师大版》

【变式】已知方程组$\begin{cases}ax - by = 4,\\2x + 3y = 4\end{cases}$与$\begin{cases}ax + by = 2,\\4x - 3y = 2\end{cases}$的解相同,那么 $a + b =$_______.
答案: $\frac{3}{2}$解析:解方程组$\begin{cases}2x + 3y = 4\\4x - 3y = 2\end{cases}$,得$\begin{cases}x = 1\\y = \frac{2}{3}\end{cases}$. 把$x$、$y$的值分别代入$ax - by = 4$、$ax + by = 2$,可得方程组$\begin{cases}a-\frac{2}{3}b = 4\\a+\frac{2}{3}b = 2\end{cases}$,解得$\begin{cases}a = 3\\b = -\frac{3}{2}\end{cases}$,$\therefore a + b = 3-\frac{3}{2}=\frac{3}{2}$.
12(易错题)解下列方程组:
(1)$\frac{2y + x}{3}=2x + 1=\frac{3y - 2}{4}$;
(2)$\begin{cases}\frac{x + 2}{6}+\frac{2y - 3}{5}=0,\\\frac{5x - 2}{6}-\frac{2y + 2}{5}=3.\end{cases}$
答案: 解:- (1)原方程组整理得$\begin{cases}2y - 5x = 3,①\\8x - 3y = - 6.②\end{cases}$①×3 + ②×2,得$x = - 3$. 把$x = - 3$代入①,得$2y + 15 = 3$,解得$y = - 6$,所以原方程组的解为$\begin{cases}x = - 3\\y = - 6\end{cases}$. - (2)$\begin{cases}\frac{x + 2}{6}+\frac{2y - 3}{5}=0,①\\\frac{5x - 2}{6}-\frac{2y + 2}{5}=3,②\end{cases}$① + ②,得$x - 1 = 3$,即$x = 4$,将$x = 4$代入①,得$1+\frac{2y - 3}{5}=0$,解得$y = - 1$,所以原方程组的解为$\begin{cases}x = 4\\y = - 1\end{cases}$. 易错点:将方程两边都乘以适当的数时,方程中的每一项(包括常数项)都要乘以这个数.
13【新趋势·阅读理解题】阅读下面解方程组的方法,然后回答问题.
解方程组$\begin{cases}19x + 18y = 17,①\\17x + 16y = 15.②\end{cases}$
解:由① - ②,得 $2x + 2y = 2$,即 $x + y = 1$③,
③×16,得 $16x + 16y = 16$④,
② - ④,得 $x = - 1$,从而可得 $y = 2$,
∴原方程组的解是$\begin{cases}x = - 1,\\y = 2.\end{cases}$
(1)请你仿照上面的解题方法解方程组$\begin{cases}23x + 22y = 21,①\\25x + 24y = 23;②\end{cases}$
(2)直接写出方程组$\begin{cases}2023x - 2024y = 2025,\\2024x - 2025y = 2026\end{cases}$的解:_______.
答案: 解:- (1)$\begin{cases}23x + 22y = 21,①\\25x + 24y = 23,②\end{cases}$由② - ①,得$2x + 2y = 2$,即$x + y = 1$③,③×24,得$24x + 24y = 24$④,② - ④,得$x = - 1$,从而可得$y = 2$,$\therefore$原方程组的解是$\begin{cases}x = - 1\\y = 2\end{cases}$.@@$\begin{cases}x = - 1\\y = - 2\end{cases}$提示:$\begin{cases}2023x - 2024y = 2025,①\\2024x - 2025y = 2026,②\end{cases}$由② - ①,得$x - y = 1$③,③×2024,得$2024x - 2024y = 2024$④,④ - ②,得$y = - 2$,从而可得$x = - 1$,$\therefore$原方程组的解是$\begin{cases}x = - 1\\y = - 2\end{cases}$. 核心素养:本题考查了核心素养的应用意识,读懂题意,根据题中方法求解即可.
1.(山西晋城期末)已知关于 $x、y$的方程组$\begin{cases}2ax - 3by = 2c,\\3ax + 2by = 16c\end{cases}$的解是$\begin{cases}x = 4,\\y = 2,\end{cases}$则关于 $x、y$的方程组$\begin{cases}2ax - 3by + 2a = 2c,\\3ax + 2by + 3a = 16c\end{cases}$的解是 ( )
A.$\begin{cases}x = 4,\\y = 2\end{cases}$
B.$\begin{cases}x = 3,\\y = 2\end{cases}$
C.$\begin{cases}x = 5,\\y = 2\end{cases}$
D.$\begin{cases}x = 5,\\y = 1\end{cases}$
答案: B 解析:关于$x$、$y$的方程组$\begin{cases}2ax - 3by + 2a = 2c\\3ax + 2by + 3a = 16c\end{cases}$可整理为$\begin{cases}2a(x + 1)-3by = 2c\\3a(x + 1)+2by = 16c\end{cases}$. $\because$关于$x$、$y$的方程组$\begin{cases}2ax - 3by = 2c\\3ax + 2by = 16c\end{cases}$的解是$\begin{cases}x = 4\\y = 2\end{cases}$,$\therefore\begin{cases}x + 1 = 4\\y = 2\end{cases}$,解得$\begin{cases}x = 3\\y = 2\end{cases}$. 故选B.
2. 解方程组:$\begin{cases}2x - 3y - 7 = 0,\\\frac{2x - 3y}{7}+2y = 9.\end{cases}$
答案: 解:$\begin{cases}2x - 3y - 7 = 0,①\\\frac{2x - 3y}{7}+2y = 9,②\end{cases}$由①,得$2x - 3y = 7$③. 将③代入②,得$1 + 2y = 9$,解得$y = 4$. 把$y = 4$代入③,得$2x - 12 = 7$,解得$x = \frac{19}{2}$,所以$\begin{cases}x = \frac{19}{2}\\y = 4\end{cases}$.

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