登录
数学六年级
1.已知算式9□(-9)的值为-1，则“□”内应填入的运算符号为（
A.+
B.−
C.×
D.÷
1.已知算式9□(-9)的值为-1，则“□”内应填入的运算符号为（
D
）A.+
B.−
C.×
D.÷
答案
9÷(-9)=-1.故选D.
2.与$10÷(-5)$结果相同的是 （
A.$\frac{1}{10}÷(-5)$
B.$\frac{1}{10}×(-5)$
C.$10×(-\frac{1}{5})$
D.$10÷(-\frac{1}{5})$
C
）A.$\frac{1}{10}÷(-5)$
B.$\frac{1}{10}×(-5)$
C.$10×(-\frac{1}{5})$
D.$10÷(-\frac{1}{5})$
答案
$10÷(-5)=10×(-\dfrac{1}{5})=-2$.故选C.
3.「2026 山东聊城东昌府期中」将$(-7)÷(-\frac{3}{4})÷(-2.5)$转化为乘法运算正确的是（
A.$(-7)×\frac{4}{3}×(-2.5)$
B.$(-7)×(-\frac{4}{3})×(-2.5)$
C.$(-7)×(-\frac{4}{3})×(-\frac{2}{5})$
D.$(-7)×(-\frac{3}{4})×(-\frac{5}{2})$
C
）A.$(-7)×\frac{4}{3}×(-2.5)$
B.$(-7)×(-\frac{4}{3})×(-2.5)$
C.$(-7)×(-\frac{4}{3})×(-\frac{2}{5})$
D.$(-7)×(-\frac{3}{4})×(-\frac{5}{2})$
答案
原式$=(-7)×(-\dfrac{4}{3})×(-\dfrac{2}{5})$,故选C.
4.学科特色
教材变式 计算:
(1)$(-3 \dfrac{2}{3})÷ 5 \dfrac{1}{2}$.
(2)$(-12)÷ \dfrac{3}{4}÷ \dfrac{2}{3}$.
(3)$(-0.5)÷ 1 \dfrac{5}{6}÷ (-\dfrac{1}{11})$.
(4)$(-9999 \dfrac{11}{13})÷ (-11)÷ (-3)$.
教材变式 计算:
(1)$(-3 \dfrac{2}{3})÷ 5 \dfrac{1}{2}$.
(2)$(-12)÷ \dfrac{3}{4}÷ \dfrac{2}{3}$.
(3)$(-0.5)÷ 1 \dfrac{5}{6}÷ (-\dfrac{1}{11})$.
(4)$(-9999 \dfrac{11}{13})÷ (-11)÷ (-3)$.
答案
(1)
$\begin{aligned}(-3\dfrac{2}{3})÷ 5\dfrac{1}{2}&=-\dfrac{11}{3}×\dfrac{2}{11}\\&=-\dfrac{2}{3}.\end{aligned}$
(2)
$\begin{aligned}(-12)÷ \dfrac{3}{4}÷ \dfrac{2}{3}&=-12×\dfrac{4}{3}×\dfrac{3}{2}\\&=-24.\end{aligned}$
(3)
$\begin{aligned}(-0.5)÷ 1\dfrac{5}{6}÷ (-\dfrac{1}{11})&=\dfrac{1}{2}×\dfrac{6}{11}×11\\&=3.\end{aligned}$
(4)
$\begin{aligned}(-9\ 999\dfrac{11}{13})÷ (-11)÷ (-3)&=(-9\ 999\dfrac{11}{13})×[(-\dfrac{1}{11})×(-\dfrac{1}{3})]\\&=(-9\ 999-\dfrac{11}{13})×\dfrac{1}{33}\\&=-9\ 999×\dfrac{1}{33}-\dfrac{11}{13}×\dfrac{1}{33}\\&=-303-\dfrac{1}{39}\\&=-303\dfrac{1}{39}.\end{aligned}$
$\begin{aligned}(-3\dfrac{2}{3})÷ 5\dfrac{1}{2}&=-\dfrac{11}{3}×\dfrac{2}{11}\\&=-\dfrac{2}{3}.\end{aligned}$
(2)
$\begin{aligned}(-12)÷ \dfrac{3}{4}÷ \dfrac{2}{3}&=-12×\dfrac{4}{3}×\dfrac{3}{2}\\&=-24.\end{aligned}$
(3)
$\begin{aligned}(-0.5)÷ 1\dfrac{5}{6}÷ (-\dfrac{1}{11})&=\dfrac{1}{2}×\dfrac{6}{11}×11\\&=3.\end{aligned}$
(4)
$\begin{aligned}(-9\ 999\dfrac{11}{13})÷ (-11)÷ (-3)&=(-9\ 999\dfrac{11}{13})×[(-\dfrac{1}{11})×(-\dfrac{1}{3})]\\&=(-9\ 999-\dfrac{11}{13})×\dfrac{1}{33}\\&=-9\ 999×\dfrac{1}{33}-\dfrac{11}{13}×\dfrac{1}{33}\\&=-303-\dfrac{1}{39}\\&=-303\dfrac{1}{39}.\end{aligned}$
5.「2026山东济南月考」计算$(-7)÷(-12)×\frac{1}{12}$的结果是（
A.$-1$
B.$1$
C.$-\frac{7}{144}$
D.$\frac{7}{144}$
D
）A.$-1$
B.$1$
C.$-\frac{7}{144}$
D.$\frac{7}{144}$
答案
$(-7)÷(-12)×\dfrac{1}{12}=\dfrac{7}{12}×\dfrac{1}{12}=\dfrac{7}{144}$.故选D.
6.若$2÷4×(-6)□8=5$,则推算出“□”内填的符号应是
+
(填“+”“-”“×”或“÷”).答案
答案 +
解析 因为$2÷4×(-6)□8=5$,所以$2×\dfrac{1}{4}×(-6)□8=5$,所以$-3□8=5$,所以可推算出“□”内填的符号应是+.
解析 因为$2÷4×(-6)□8=5$,所以$2×\dfrac{1}{4}×(-6)□8=5$,所以$-3□8=5$,所以可推算出“□”内填的符号应是+.
7.计算：
(1) $-3\dfrac{1}{3}÷(-\dfrac{4}{9})×1\dfrac{1}{3}$.
(2) $-2.5÷\dfrac{5}{32}×(-\dfrac{1}{8})÷(-4)$.
(1) $-3\dfrac{1}{3}÷(-\dfrac{4}{9})×1\dfrac{1}{3}$.
(2) $-2.5÷\dfrac{5}{32}×(-\dfrac{1}{8})÷(-4)$.
答案
(1)
$\begin{aligned}-3\dfrac{1}{3}÷(-\dfrac{4}{9})×1\dfrac{1}{3}&=-\dfrac{10}{3}×(-\dfrac{9}{4})×\dfrac{4}{3}\\&=10.\end{aligned}$
(2)
$\begin{aligned}-2.5÷\dfrac{5}{32}×(-\dfrac{1}{8})÷(-4)&=-\dfrac{5}{2}×\dfrac{32}{5}×(-\dfrac{1}{8})×(-\dfrac{1}{4})\\&=-\dfrac{5}{2}×\dfrac{32}{5}×\dfrac{1}{8}×\dfrac{1}{4}\\&=-\dfrac{1}{2}.\end{aligned}$
$\begin{aligned}-3\dfrac{1}{3}÷(-\dfrac{4}{9})×1\dfrac{1}{3}&=-\dfrac{10}{3}×(-\dfrac{9}{4})×\dfrac{4}{3}\\&=10.\end{aligned}$
(2)
$\begin{aligned}-2.5÷\dfrac{5}{32}×(-\dfrac{1}{8})÷(-4)&=-\dfrac{5}{2}×\dfrac{32}{5}×(-\dfrac{1}{8})×(-\dfrac{1}{4})\\&=-\dfrac{5}{2}×\dfrac{32}{5}×\dfrac{1}{8}×\dfrac{1}{4}\\&=-\dfrac{1}{2}.\end{aligned}$
8.「2025山东德州月考，★☆」计算$\frac{1}{6}×(-6)÷(-\frac{1}{6})×6$的结果是（
A.6
B.36
C.-1
D.1
B
）A.6
B.36
C.-1
D.1
答案
$\dfrac{1}{6}×(-6)÷(-\dfrac{1}{6})×6=\dfrac{1}{6}×(-6)×(-6)×6=36$.故选B.
9.「2026河北邯郸期中，★☆」文文在计算$(-24)÷ a$时，误将“÷”看成“+”，结果是-8，则$-16÷ a$的正确结果是
-1
。答案
答案 -1
解析 根据题意可知,$a=-8-(-24)=-8+24=16$,
$\therefore -16÷ a=-16÷16=-1$.
故答案为-1.
解析 根据题意可知,$a=-8-(-24)=-8+24=16$,
$\therefore -16÷ a=-16÷16=-1$.
故答案为-1.
10.「★☆」在-2,-3,0,4这四个数中,任意两个数相除,所得的商最小是
-2
.答案
答案 -2
解析 取异号且商的绝对值最大的两数相除,可得商最小是$4÷(-2)=-2$.
解析 取异号且商的绝对值最大的两数相除,可得商最小是$4÷(-2)=-2$.
11.「2025山东临沂兰山月考，★☆」有两个数-4和+6，它们相反数的和为$a$，倒数的和为$b$，和的倒数为$c$，求$a÷b÷c$的值.
答案
因为-4的相反数是4,+6的相反数是-6,-4的倒数是$-\dfrac{1}{4}$,+6的倒数是$\dfrac{1}{6}$,$-4+6=2$,
所以$a=4+(-6)=-2$,$b=-\dfrac{1}{4}+\dfrac{1}{6}=-\dfrac{1}{12}$,$c=\dfrac{1}{2}$,
所以$a÷ b÷ c=-2÷(-\dfrac{1}{12})÷\dfrac{1}{2}=2×12×2=48$.
所以$a=4+(-6)=-2$,$b=-\dfrac{1}{4}+\dfrac{1}{6}=-\dfrac{1}{12}$,$c=\dfrac{1}{2}$,
所以$a÷ b÷ c=-2÷(-\dfrac{1}{12})÷\dfrac{1}{2}=2×12×2=48$.
