《2025年暑假作业新疆青少年出版社七年级数学人教版》

1. 下列选项中是一元一次不等式组的是 ()
A. $\left\{\begin{array}{l} x-y＞0,\\ y+z＞0.\end{array}\right. $
B. $\left\{\begin{array}{l} x^{2}-x＞0,\\ x+1＜0.\end{array}\right. $
C. $\left\{\begin{array}{l} y+2＞0,\\ x+y＜0.\end{array}\right. $
D. $\left\{\begin{array}{l} 2x+3＞0,\\ x＞0.\end{array}\right. $

答案： D

2. 在下列不等式组中,解集为$-1≤x＜4$的是 ()
A. $\left\{\begin{array}{l} x≥-1,\\ 4＜x.\end{array}\right. $
B. $\left\{\begin{array}{l} x≥-1,\\ x＞4.\end{array}\right. $
C. $\left\{\begin{array}{l} x-4＜0,\\ x+1≥0.\end{array}\right. $
D. $\left\{\begin{array}{l} x-4＞0,\\ x≥-1.\end{array}\right. $

答案： C

3. 不等式组$\left\{\begin{array}{l} x+2＞0,\\ 2x-6≤0\end{array}\right. $

的解集在数轴上表示正确的是 ()

答案： C

4. 关于$x的不等式组\left\{\begin{array}{l} x-m＞0,\\ 2x-3≥3(x-2)\end{array}\right. $恰有四个整数解,那么$m$的取值范围为 ()
A. $m≥-1$
B. $m＜0$
C. $-1≤m＜0$
D. $-1＜m＜0$

答案： C

5. 如果关于$x的方程ax-2= x+3$的解为非正数,且关于$x,y的二元一次方程组\left\{\begin{array}{l} x+3y= 3,\\ 3x+y= 1+a\end{array}\right. 的解满足x+y＞-\frac {3}{4}$,则满足条件的整数$a$有 ()
A. 5个
B. 6个
C. 7个
D. 8个

答案： C

6. 定义$[x]为不超过x$的最大整数,如$[3.6]= 3,[0.6]= 0,[-3.6]= -4$.对于任意实数$x$,下列式子中错误的是 ()
A. $[x]= x$($x$为整数)
B. $0≤x-[x]＜1$
C. $[x+y]≤[x]+[y]$
D. $[n+x]= n+[x]$($n$为整数)

答案： C

7. 现在有住宿生若干名,分住若干间宿舍,若每间住4人,则还有19人无宿舍住;若每间住6人,则有一间宿舍不空也不满,若设宿舍间数为$x$,则可以列得不等式组为 ()
A. $\left\{\begin{array}{l} (4x+19)-6(x-1)≥1,\\ (4x+19)-6(x-1)≤6.\end{array}\right. $
B. $\left\{\begin{array}{l} (4x+19)-6(x-1)≤1,\\ (4x+19)-6(x-1)≥6.\end{array}\right. $
C. $\left\{\begin{array}{l} (4x+19)-6(x-1)≤1,\\ (4x+19)-6(x-1)≥5.\end{array}\right. $
D. $\left\{\begin{array}{l} (4x+19)-6(x-1)≥1,\\ (4x+19)-6(x-1)≤5.\end{array}\right. $

答案： D

17. 已知关于$x,y的方程组\left\{\begin{array}{l} x-2y= m\quad①,\\ 2x+3y= 2m+4\quad②\end{array}\right. 的解满足不等式组\left\{\begin{array}{l} 3x+y≤0,\\ x+5y＞0,\end{array}\right. 求满足条件的m$的整数值.

答案： ①×2得：$2x - 4y = 2m$ ③，② - ③得：$y = \frac{4}{7}$，把$y = \frac{4}{7}$代入①得：$x = m + \frac{8}{7}$，把$x = m + \frac{8}{7}$，$y = \frac{4}{7}$代入不等式组$\begin{cases} 3x + y \leq 0 \\ x + 5y ＞ 0 \end{cases}$中得：$\begin{cases} 3m + 4 \leq 0 \\ m + 4 ＞ 0 \end{cases}$，解不等式组得：$-4 ＜ m \leq -\frac{4}{3}$，
∵m为整数，
∴$m = -3$或$m = -2$.