1. 简化计算$(\dfrac{1}{24}-\dfrac{7}{12}+\dfrac{1}{4})× (-24)$,应该运用( )。

A.加法交换律
B.加法结合律
C.乘法对加法的分配律
D.乘法结合律

C

2. 对式子$(-\dfrac{3}{5}+\dfrac{1}{2})× (-4)× (-25)$进行简便计算,如图所示,①运用的运算律是( )。
$\begin{aligned}解：原式&= (-\dfrac{3}{5}+\dfrac{1}{2})× [(-4)× (-25)]…(①)\\&=(-\dfrac{3}{5}+\dfrac{1}{2})× 100\\&=-\dfrac{3}{5}× 100 + \dfrac{1}{2}× 100\\&= -60 + 50\\&= -10。\end{aligned} $

A.乘法交换律
B.乘法对加法的分配律
C.乘法结合律
D.加法交换律

C

3. 用简便方法计算：$47× (-\dfrac{1}{8}) + 81×\dfrac{1}{8} + 26×(-0.125)$,其结果是( )。

A.2
B.1
C.0
D.-1

B

$47×\left(-\dfrac{1}{8}\right)+81×\dfrac{1}{8}+26×\left(-0.125\right)$
$=47×\left(-\dfrac{1}{8}\right)+81×\dfrac{1}{8}+26×\left(-\dfrac{1}{8}\right)$
$=\left(-47+81-26\right)×\dfrac{1}{8}$
$=8×\dfrac{1}{8}$
$=1$
B

4. 计算$-6× \dfrac{2}{3}× \vert -\dfrac{1}{6}\vert × 1\dfrac{1}{2}$的值为( )。

A.1
B.36
C.-1
D.0

C

$-6× \dfrac{2}{3}× \left\vert -\dfrac{1}{6}\right\vert × 1\dfrac{1}{2}$
$=-6× \dfrac{2}{3}× \dfrac{1}{6}× \dfrac{3}{2}$
$=(-6× \dfrac{1}{6})× (\dfrac{2}{3}× \dfrac{3}{2})$
$=-1× 1$
$=-1$
C

5. 利用乘法运算律填空：
(1)$(-1.25)× 3× (-8) = (-1.25)× $_________$×3$；
(2)$[2× (-4)]× (+\dfrac{1}{4}) = 2×[$_________$]$；
(3)$27× (-1\dfrac{1}{9}) = 27×$_________$ + 27×$_________$$。

（1）$(-8)$（2）$(-4)×\left(+\dfrac{1}{4}\right)$（3）$(-1)\left(-\dfrac{1}{9}\right)$

6. 计算：
(1)$(\dfrac{5}{3}-\dfrac{3}{4}-\dfrac{5}{12})× (-4)$；
(2)$(-2)× (-7)× 5× (-\dfrac{1}{7})$；
(3)$(-3)× (-\dfrac{7}{5})× (-\dfrac{1}{3})× \dfrac{4}{7}$；
(4)$2.25×4.8 + 77.5×48\%$；
(5)$(-\dfrac{1}{6}+\dfrac{3}{4}-\dfrac{1}{12})× 48$。

（1）$-2$（2）$-10$（3）$-\dfrac{4}{5}$（4）48（5）24

(1)
$\begin{aligned}\left(\dfrac{5}{3}-\dfrac{3}{4}-\dfrac{5}{12}\right)× (-4)&=\dfrac{5}{3}× (-4)-\dfrac{3}{4}× (-4)-\dfrac{5}{12}× (-4)\\&=-\dfrac{20}{3}+3+\dfrac{5}{3}\\&=-\dfrac{15}{3}+3\\&=-5 + 3\\&=-2\end{aligned}$
(2)
$\begin{aligned}(-2)× (-7)× 5× \left(-\dfrac{1}{7}\right)&=(-2)× 5× [(-7)× \left(-\dfrac{1}{7}\right)]\\&=-10× 1\\&=-10\end{aligned}$
(3)
$\begin{aligned}(-3)× \left(-\dfrac{7}{5}\right)× \left(-\dfrac{1}{3}\right)× \dfrac{4}{7}&=(-3)× \left(-\dfrac{1}{3}\right)× \left(-\dfrac{7}{5}\right)× \dfrac{4}{7}\\&=1× \left(-\dfrac{4}{5}\right)\\&=-\dfrac{4}{5}\end{aligned}$
(4)
$\begin{aligned}2.25× 4.8 + 77.5× 48\%&=2.25× 4.8 + 77.5× 0.48\\&=2.25× 4.8 + 7.75× 4.8\\&=(2.25 + 7.75)× 4.8\\&=10× 4.8\\&=48\end{aligned}$
(5)
$\begin{aligned}\left(-\dfrac{1}{6}+\dfrac{3}{4}-\dfrac{1}{12}\right)× 48&=-\dfrac{1}{6}× 48+\dfrac{3}{4}× 48-\dfrac{1}{12}× 48\\&=-8 + 36 - 4\\&=24\end{aligned}$

7. 若“!”是一种数学运算符号,并且$1! = 1$,$2! = 2×1 = 2$,$3! = 3×2×1 = 6$,$4! = 4×3×2×1$,…$$,则$\dfrac{100!}{98!}= $_________$$。

9900

$\dfrac{100!}{98!}=\dfrac{100×99×98!}{98!}=100×99=9900$