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1.「2026广西梧州期末」计算$(-108)×(-25)×4$的结果是（
A.10 800
B.-2 700
C.-432
D.1 080
A
）A.10 800
B.-2 700
C.-432
D.1 080
答案
$(-108)×(-25)×4=+(108×25×4)=10 800.$故选A.
2.「2026河南驻马店期中」绝对值小于5的所有整数的积是（
A.-10
B.0
C.10
D.20
B
）A.-10
B.0
C.10
D.20
答案
绝对值小于5的所有整数为-4,-3,-2,-1,0,1,2,3,4,$-4×(-3)×(-2)×(-1)×0×1×2×3×4=0.$故选B.
3.计算:
(1)$(-\dfrac{3}{5})×8×(-\dfrac{4}{3})$.
(2)$2.25×(-3.5)×(-40)×20$.
(3)$(-\dfrac{5}{12})× $
$ ×1.5×(-1\dfrac{1}{4})$.
(1)$(-\dfrac{3}{5})×8×(-\dfrac{4}{3})$.
(2)$2.25×(-3.5)×(-40)×20$.
(3)$(-\dfrac{5}{12})× $
$ ×1.5×(-1\dfrac{1}{4})$.答案
(1)$(-\dfrac{3}{5})×8×(-\dfrac{4}{3})=+(\dfrac{3}{5}×8×\dfrac{4}{3})=\dfrac{32}{5}.$
(2)$2.25×(-3.5)×(-40)×20=+(2.25×3.5×40×20)=6 300.$
(3)$(-\dfrac{5}{12})×\dfrac{4}{15}×1.5×(-1\dfrac{1}{4})=\dfrac{5}{12}×\dfrac{4}{15}×\dfrac{3}{2}×\dfrac{5}{4}=\dfrac{5}{24}.$
(2)$2.25×(-3.5)×(-40)×20=+(2.25×3.5×40×20)=6 300.$
(3)$(-\dfrac{5}{12})×\dfrac{4}{15}×1.5×(-1\dfrac{1}{4})=\dfrac{5}{12}×\dfrac{4}{15}×\dfrac{3}{2}×\dfrac{5}{4}=\dfrac{5}{24}.$
4.「2026安徽黄山月考」下列各式中，运用运算律不正确的是（
A.$(-4)×3=3×(-4)$
B.$-24×\frac{1}{5}×(-\frac{1}{6})=\frac{1}{5}×[(-24)×(-\frac{1}{6})]$
C.$(-5)×\frac{1}{2}×(-6)=(-5)×[\frac{1}{2}×(-6)]$
D.$(-12)×[\frac{1}{3}+(-\frac{1}{4})]=(-12)×\frac{1}{3}+(-\frac{1}{4})$
D
）A.$(-4)×3=3×(-4)$
B.$-24×\frac{1}{5}×(-\frac{1}{6})=\frac{1}{5}×[(-24)×(-\frac{1}{6})]$
C.$(-5)×\frac{1}{2}×(-6)=(-5)×[\frac{1}{2}×(-6)]$
D.$(-12)×[\frac{1}{3}+(-\frac{1}{4})]=(-12)×\frac{1}{3}+(-\frac{1}{4})$
答案
A,B,C三项的计算正确;D项,$(-12)×[\dfrac{1}{3}+(-\dfrac{1}{4})]=(-12)×\dfrac{1}{3}+(-12)×(-\dfrac{1}{4})$,故原计算错误.故选D.
5.「2026天津河西月考」要使$(\frac{3}{11} × \frac{6}{7}) × 11 × 7$计算简便,可运用 （
A.乘法交换律
B.乘法结合律
C.乘法交换律和结合律
D.乘法分配律
C
）A.乘法交换律
B.乘法结合律
C.乘法交换律和结合律
D.乘法分配律
答案
$(\dfrac{3}{11}×\dfrac{6}{7})×11×7=\dfrac{3}{11}×\dfrac{6}{7}×11×7=\dfrac{3}{11}×11×\dfrac{6}{7}×7=(\dfrac{3}{11}×11)×(\dfrac{6}{7}×7)$,所以要使计算简便,可运用乘法交换律和结合律.故选C.
6. 学科特色易错题「2025河北中考」一道习题及其错误的解答过程如下：
请指出在第几步开始出现错误，并选择你喜欢的方法写出正确的解答过程。
请指出在第几步开始出现错误，并选择你喜欢的方法写出正确的解答过程。
答案
在第一步开始出现错误,正确解答过程如下：
$(-6)×(\dfrac{1}{2}+\dfrac{2}{3}-\dfrac{5}{6})$
$=(-6)×\dfrac{1}{2}+(-6)×\dfrac{2}{3}-(-6)×\dfrac{5}{6}$
$=-3-4+5$
$=-2.$
$(-6)×(\dfrac{1}{2}+\dfrac{2}{3}-\dfrac{5}{6})$
$=(-6)×\dfrac{1}{2}+(-6)×\dfrac{2}{3}-(-6)×\dfrac{5}{6}$
$=-3-4+5$
$=-2.$
7.计算:(1)$1.25×(-\dfrac{8}{25})×(-8).$
(2)$(-3\dfrac{1}{5})×(-7\dfrac{2}{7})×\dfrac{21}{51}×\dfrac{25}{16}.$
(3)$(-\dfrac{1}{6}+\dfrac{3}{4}-\dfrac{1}{12})×(-48).$
(4)$-100×(\dfrac{3}{10}-\dfrac{1}{2}+\dfrac{1}{5}-0.1).$
(2)$(-3\dfrac{1}{5})×(-7\dfrac{2}{7})×\dfrac{21}{51}×\dfrac{25}{16}.$
(3)$(-\dfrac{1}{6}+\dfrac{3}{4}-\dfrac{1}{12})×(-48).$
(4)$-100×(\dfrac{3}{10}-\dfrac{1}{2}+\dfrac{1}{5}-0.1).$
答案
(1)$1.25×(-\dfrac{8}{25})×(-8)$
$=[1.25×(-8)]×(-\dfrac{8}{25})$
$=(-10)×(-\dfrac{8}{25})=\dfrac{16}{5}.$
(2)$(-3\dfrac{1}{5})×(-7\dfrac{2}{7})×\dfrac{21}{51}×\dfrac{25}{16}$
$=\dfrac{16}{5}×\dfrac{25}{16}×\dfrac{51}{7}×\dfrac{21}{51}$
$=5×3=15.$
(3)$(-\dfrac{1}{6}+\dfrac{3}{4}-\dfrac{1}{12})×(-48)$
$=(-\dfrac{1}{6})×(-48)+\dfrac{3}{4}×(-48)-\dfrac{1}{12}×(-48)$
$=8-36+4=-24.$
(4)$-100×(\dfrac{3}{10}-\dfrac{1}{2}+\dfrac{1}{5}-0.1)$
$=-100×\dfrac{3}{10}+100×\dfrac{1}{2}-100×\dfrac{1}{5}+100×0.1$
$=-30+50-20+10=10.$
$=[1.25×(-8)]×(-\dfrac{8}{25})$
$=(-10)×(-\dfrac{8}{25})=\dfrac{16}{5}.$
(2)$(-3\dfrac{1}{5})×(-7\dfrac{2}{7})×\dfrac{21}{51}×\dfrac{25}{16}$
$=\dfrac{16}{5}×\dfrac{25}{16}×\dfrac{51}{7}×\dfrac{21}{51}$
$=5×3=15.$
(3)$(-\dfrac{1}{6}+\dfrac{3}{4}-\dfrac{1}{12})×(-48)$
$=(-\dfrac{1}{6})×(-48)+\dfrac{3}{4}×(-48)-\dfrac{1}{12}×(-48)$
$=8-36+4=-24.$
(4)$-100×(\dfrac{3}{10}-\dfrac{1}{2}+\dfrac{1}{5}-0.1)$
$=-100×\dfrac{3}{10}+100×\dfrac{1}{2}-100×\dfrac{1}{5}+100×0.1$
$=-30+50-20+10=10.$
