12.「2025 山东德州月考改编，★☆」阅读下面解题过程并解答问题.
计算：$(-15)÷(-\dfrac{1}{2}×\dfrac{25}{3})÷\dfrac{1}{6}$.
解：原式$=(-15)÷(-\dfrac{25}{6})×6$(第一步)
$=(-15)÷(-25)$(第二步)
$=-\dfrac{3}{5}$.(第三步)
(1)上面解题过程有两处错误：
第一处是第步，错误原因是；
第二处是第步，错误原因是.
(2)请计算出正确的结果.

(1)二;没有按同级运算从左至右运算;三;正负符号错误.
(2)
$\begin{aligned}(-15)÷(-\dfrac{1}{2}×\dfrac{25}{3})÷\dfrac{1}{6}&=-15÷(-\dfrac{25}{6})×6\\&=15×\dfrac{6}{25}×6=\dfrac{108}{5}.\end{aligned}$

13. 核心素养 运算能力 阅读材料：
计算：$(-\dfrac{1}{30})÷(\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5})$.
解法一：$(-\dfrac{1}{30})÷(\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5})$
$=(-\dfrac{1}{30})÷\dfrac{2}{3}-(-\dfrac{1}{30})÷\dfrac{1}{10}+(-\dfrac{1}{30})÷\dfrac{1}{6}-(-\dfrac{1}{30})÷\dfrac{2}{5}$
$=-\dfrac{1}{20}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{12}=\dfrac{1}{6}$.
解法二：$(-\dfrac{1}{30})÷(\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5})$
$=(-\dfrac{1}{30})÷[(\dfrac{2}{3}+\dfrac{1}{6})-(\dfrac{1}{10}+\dfrac{2}{5})]$
$=(-\dfrac{1}{30})÷(\dfrac{5}{6}-\dfrac{1}{2})=-\dfrac{1}{30}×3=-\dfrac{1}{10}$.
解法三：$\because(\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5})÷(-\dfrac{1}{30})$
$=(\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5})×(-30)$
$=-20+3-5+12=-10$，$\therefore$原式$=-\dfrac{1}{10}$.
解决问题：
(1) 上述三种解法得出的结果不同，你认为解法几是错误的？
(2) 在正确的解法中，你认为解法几最简便？
(3) 计算：$(-\dfrac{1}{42})÷(\dfrac{1}{6}-\dfrac{3}{14}+\dfrac{2}{3}-\dfrac{2}{7})$.

(1)解法一是错误的.
(2)正确的解法中解法三最简便.
(3)【解法一】
$\begin{aligned}(-\dfrac{1}{42})÷(\dfrac{1}{6}-\dfrac{3}{14}+\dfrac{2}{3}-\dfrac{2}{7})&=(-\dfrac{1}{42})÷[(\dfrac{1}{6}+\dfrac{2}{3})-(\dfrac{3}{14}+\dfrac{2}{7})]\\&=(-\dfrac{1}{42})÷(\dfrac{5}{6}-\dfrac{1}{2})\\&=(-\dfrac{1}{42})×3\\&=-\dfrac{1}{14}.\end{aligned}$
【解法二】
$\because (\dfrac{1}{6}-\dfrac{3}{14}+\dfrac{2}{3}-\dfrac{2}{7})÷(-\dfrac{1}{42})$
$=(\dfrac{1}{6}-\dfrac{3}{14}+\dfrac{2}{3}-\dfrac{2}{7})×(-42)$
$=-7+9-28+12=-14$,
$\therefore (-\dfrac{1}{42})÷(\dfrac{1}{6}-\dfrac{3}{14}+\dfrac{2}{3}-\dfrac{2}{7})=-\dfrac{1}{14}.$

1.「2026山东聊城期中」有理数$a,b$在数轴上的对应点的位置如图所示，则下面式子:①$b＜0＜a$；②$|b|＜|a|$；③$ab＞0$；④$a+b＜0$.其中正确的是 ()


A.①②
B.②③④
C.②④
D.①②③④

1.C 根据题中数轴可知$a＜0＜b,|a|＞|b|$,
所以$ab＜0,a+b＜0$,
所以②④正确,①③错误.故选 C.

2.「2026山东青岛月考」若$|m|=2$，$|n|=3$，且$|m+n|=m+n$，则$\dfrac{n}{m}=$（）

A.$\dfrac{3}{2}$
B.$-\dfrac{3}{2}$
C.$\dfrac{3}{2}$或$-\dfrac{3}{2}$
D.$\dfrac{2}{3}$或$-\dfrac{2}{3}$

2.C $\because |m|=2,|n|=3$,
$\therefore m=\pm2,n=\pm3.$
$\because |m+n|=m+n$,
$\therefore m=2,n=3$ 或 $m=-2,n=3.$
当$m=2,n=3$ 时,$\dfrac{n}{m}=\dfrac{3}{2}$;
当$m=-2,n=3$ 时,$\dfrac{n}{m}=\dfrac{3}{-2}=-\dfrac{3}{2}$.故选 C.

3.计算$32÷(-4)×\frac{1}{4}$的结果是.

3.答案 -2
解析 $32÷(-4)×\dfrac{1}{4}=32×(-\dfrac{1}{4})×\dfrac{1}{4}=-2.$