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10.(2024年吉林省改编)等式$[(-8)-□]÷(-2)=4$中，$□$表示的数是 (
A.1
B.−1
C.−2
D.0
D
)A.1
B.−1
C.−2
D.0
答案
D
11.一架直升机从高度为500米的位置开始，先以20米/秒的速度垂直上升60秒，后以12米/秒的速度垂直下降100秒，这时飞机所在的高度是
500
米。答案
500
12.在如图所示的运算流程中，若输入的数为3，则输出的数为
-2
。答案
-2
13. 计算：
(1) $0 ÷ (-2024\dfrac{1}{2024})$；
(2) $-3\dfrac{2}{3} ÷ (-5\dfrac{1}{2})$；
(3) $(-0.75) ÷ \dfrac{1}{4} ÷ (-0.3)$；
(4) $(1+\dfrac{1}{3}) ÷ (\dfrac{1}{3}-1) × \dfrac{3}{8} -1$。
(1) $0 ÷ (-2024\dfrac{1}{2024})$；
(2) $-3\dfrac{2}{3} ÷ (-5\dfrac{1}{2})$；
(3) $(-0.75) ÷ \dfrac{1}{4} ÷ (-0.3)$；
(4) $(1+\dfrac{1}{3}) ÷ (\dfrac{1}{3}-1) × \dfrac{3}{8} -1$。
答案
(1) 解:原式=0.
(2) 解:原式=$(-\dfrac{11}{3})×(-\dfrac{2}{11})=\dfrac{2}{3}$.
(3) 解:原式=$-\dfrac{3}{4}÷\dfrac{1}{4}÷(-\dfrac{3}{10})=-\dfrac{3}{4}×4×(-\dfrac{10}{3})=10$.
(4) 解:原式=$\dfrac{4}{3}÷(-\dfrac{2}{3})×\dfrac{3}{8}-1=\dfrac{4}{3}×(-\dfrac{3}{2})×\dfrac{3}{8}-1=-\dfrac{3}{4}-1=-1\dfrac{3}{4}$.
(2) 解:原式=$(-\dfrac{11}{3})×(-\dfrac{2}{11})=\dfrac{2}{3}$.
(3) 解:原式=$-\dfrac{3}{4}÷\dfrac{1}{4}÷(-\dfrac{3}{10})=-\dfrac{3}{4}×4×(-\dfrac{10}{3})=10$.
(4) 解:原式=$\dfrac{4}{3}÷(-\dfrac{2}{3})×\dfrac{3}{8}-1=\dfrac{4}{3}×(-\dfrac{3}{2})×\dfrac{3}{8}-1=-\dfrac{3}{4}-1=-1\dfrac{3}{4}$.
14.(核心素养·运算能力)阅读下列材料：
计算：$\frac{1}{24} ÷ ( \frac{1}{3} - \frac{1}{4} + \frac{1}{12} )$.
解法一：原式$=\frac{1}{24} ÷ \frac{1}{3} - \frac{1}{24} ÷ \frac{1}{4} + \frac{1}{24} ÷ \frac{1}{12}$
$=\frac{1}{24} × 3 - \frac{1}{24} × 4 + \frac{1}{24} × 12 = \frac{11}{24}$.
解法二：原式$=\frac{1}{24} ÷ ( \frac{4}{12} - \frac{3}{12} + \frac{1}{12} ) = \frac{1}{24} ÷ \frac{2}{12} = \frac{1}{24} × 6 = \frac{1}{4}$.
解法三：原式的倒数$= ( \frac{1}{3} - \frac{1}{4} + \frac{1}{12} ) ÷ \frac{1}{24}$
$= ( \frac{1}{3} - \frac{1}{4} + \frac{1}{12} ) × 24 = \frac{1}{3} × 24 - \frac{1}{4} × 24 + \frac{1}{12} × 24 = 4$.
所以原式$= \frac{1}{4}$.
(1)上述得到的结果不同，你认为解法________是错误的；
(2)请你选择合适的解法计算：$( -\frac{1}{42} ) ÷ ( \frac{1}{6} - \frac{3}{14} + \frac{2}{3} - \frac{2}{7} )$.
计算：$\frac{1}{24} ÷ ( \frac{1}{3} - \frac{1}{4} + \frac{1}{12} )$.
解法一：原式$=\frac{1}{24} ÷ \frac{1}{3} - \frac{1}{24} ÷ \frac{1}{4} + \frac{1}{24} ÷ \frac{1}{12}$
$=\frac{1}{24} × 3 - \frac{1}{24} × 4 + \frac{1}{24} × 12 = \frac{11}{24}$.
解法二：原式$=\frac{1}{24} ÷ ( \frac{4}{12} - \frac{3}{12} + \frac{1}{12} ) = \frac{1}{24} ÷ \frac{2}{12} = \frac{1}{24} × 6 = \frac{1}{4}$.
解法三：原式的倒数$= ( \frac{1}{3} - \frac{1}{4} + \frac{1}{12} ) ÷ \frac{1}{24}$
$= ( \frac{1}{3} - \frac{1}{4} + \frac{1}{12} ) × 24 = \frac{1}{3} × 24 - \frac{1}{4} × 24 + \frac{1}{12} × 24 = 4$.
所以原式$= \frac{1}{4}$.
(1)上述得到的结果不同，你认为解法________是错误的；
(2)请你选择合适的解法计算：$( -\frac{1}{42} ) ÷ ( \frac{1}{6} - \frac{3}{14} + \frac{2}{3} - \frac{2}{7} )$.
答案
(1) 一
(2) 解:原式的倒数为$(\dfrac{1}{6}-\dfrac{3}{14}+\dfrac{2}{3}-\dfrac{2}{7})÷(-\dfrac{1}{42})=(\dfrac{1}{6}-\dfrac{3}{14}+\dfrac{2}{3}-\dfrac{2}{7})×(-42)=-7+9-28+12=-14$.所以原式=$-\dfrac{1}{14}$.
(2) 解:原式的倒数为$(\dfrac{1}{6}-\dfrac{3}{14}+\dfrac{2}{3}-\dfrac{2}{7})÷(-\dfrac{1}{42})=(\dfrac{1}{6}-\dfrac{3}{14}+\dfrac{2}{3}-\dfrac{2}{7})×(-42)=-7+9-28+12=-14$.所以原式=$-\dfrac{1}{14}$.
