《2025年评优监测课时作业七年级数学下册苏科版》

1. 已知x，y是整数，满足x-y+3=0，ax-y-a=0，则整数a的所有可能值有（ ）
A. 4个 B. 5个 C. 6个 D. 8个

答案： C
解析：由x-y+3=0得y=x+3，代入ax-y-a=0得ax-(x+3)-a=0→(a-1)x=a+3→x=(a+3)/(a-1)=1+4/(a-1)．
∵x为整数，
∴4/(a-1)为整数，a-1是4的因数：±1，±2，±4，
∴a-1=1→a=2；a-1=-1→a=0；a-1=2→a=3；a-1=-2→a=-1；a-1=4→a=5；a-1=-4→a=-3，共6个，选C．

4. 用合适的方法解下列方程组.
(1)$\left\{\begin{array}{l} 4(\frac {2}{15}x-y)-3(\frac {1}{15}x+y)=6,\\ 4(\frac {2}{15}x-y)+3(\frac {1}{15}x+y)=18;\end{array}\right. $
(2)$\left\{\begin{array}{l} \frac {1}{6}(4x-7y)+\frac {1}{11}(4x+7y)=\frac {1}{2},\\ \frac {1}{6}(4x-7y)-\frac {1}{22}(4x+7y)=-1.\end{array}\right. $

答案：
(1)$\left\{\begin{array}{l} x=25\\ y=\frac{1}{3}\end{array}\right. $
设$u=4(\frac{2}{15}x-y)$，$v=3(\frac{1}{15}x+y)$，则$\left\{\begin{array}{l} u-v=6\\ u+v=18\end{array}\right. $，解得u=12，v=6。则$\left\{\begin{array}{l} 4(\frac{2}{15}x-y)=12\\ 3(\frac{1}{15}x+y)=6\end{array}\right. $，化简得$\left\{\begin{array}{l} \frac{2}{15}x-y=3\\ \frac{1}{15}x+y=2\end{array}\right. $，两式相加得$\frac{3}{15}x=5$，x=25，代入得y=$\frac{1}{3}$。
(2)$\left\{\begin{array}{l} x=1\\ y=1\end{array}\right. $
设$m=\frac{1}{6}(4x-7y)$，$n=\frac{1}{11}(4x+7y)$，则$\left\{\begin{array}{l} m+n=\frac{1}{2}\\ m-\frac{n}{2}=-1\end{array}\right. $，解得m=$-\frac{1}{2}$，n=1。则$\left\{\begin{array}{l} \frac{1}{6}(4x-7y)=-\frac{1}{2}\\ \frac{1}{11}(4x+7y)=1\end{array}\right. $，化简得$\left\{\begin{array}{l} 4x-7y=-3\\ 4x+7y=11\end{array}\right. $，两式相加得8x=8，x=1，代入得y=1。