《2025年时习之暑假衔接七年级数学人教版》

(1)计算：$\sqrt {16}-(-2)^{2}-\sqrt [3]{-2^{3}}$；
(2)计算：$-2^{2}+\sqrt [3]{\frac {27}{8}}+\sqrt {1.44}-| \sqrt {(-2)^{2}}-\sqrt {5}| $；
(3)解方程组：$\left\{\begin{array}{l} x+y= 10,\\ 2x+y= 16;\end{array} \right. $
(4)解方程组：$\left\{\begin{array}{l} \frac {x}{4}+\frac {y}{3}= \frac {4}{3},\\ 5(x-9)= 6(y-2);\end{array} \right. $
(5)解不等式：$2-\frac {2x-1}{6}＞\frac {2+x}{3}$；
(6)解不等式组：$\left\{\begin{array}{l} x-3(x-2)\geqslant 4,\\ \frac {2x-1}{5}\geqslant \frac {x+1}{2}.\end{array} \right. $

答案：
(1)原式=2.
(2)原式=$0.7-\sqrt{5}$.
(3)$\begin{cases}x=6,\\y=4.\end{cases}$
(4)$\begin{cases}x=6,\\y=-\dfrac{1}{2}.\end{cases}$
(5)$x＜\dfrac{9}{4}$.
(6)$x\leqslant-7$.

1.不等式组$\left\{\begin{array}{l} x+1\geqslant 0,\\ x-2\leqslant 0\end{array} \right. $的解集是( )

A.$x\geqslant -1$
B.$x\leqslant 2$
C.$1\leqslant x\leqslant 2$
D.$-1\leqslant x\leqslant 2$

答案： D

2.不等式组$\left\{\begin{array}{l} 3x\geqslant 9,\\ x＜5\end{array} \right. $的整数解共有( )

A.1个
B.2个
C.3个
D.4个

答案： B