1. 填空：
(1) $(-\frac{2}{3})×6=$. (2) $\frac{2}{3}×(-6)=$.
(3) $(-12)×(+\frac{1}{3})×(-\frac{1}{6})=$. (4) $\frac{1}{2}×\frac{4}{3}+\frac{1}{2}×\frac{2}{3}=$.

(1)-4；(2)-4；(3)$\frac{2}{3}$；(4)1

(1) $(-\frac{2}{3})×6 = -(\frac{2}{3}×6) = -4$
(2) $\frac{2}{3}×(-6) = -(\frac{2}{3}×6) = -4$
(3) $(-12)×(+\frac{1}{3})×(-\frac{1}{6}) = [(-12)×\frac{1}{3}]×(-\frac{1}{6}) = (-4)×(-\frac{1}{6}) = \frac{4}{6} = \frac{2}{3}$
(4) $\frac{1}{2}×\frac{4}{3}+\frac{1}{2}×\frac{2}{3} = \frac{1}{2}×(\frac{4}{3}+\frac{2}{3}) = \frac{1}{2}×2 = 1$

2. 几个不等于 0 的有理数相乘,它们的积的符号()
A.由正因数的个数决定
B.由因数的个数决定
C.由负因数的个数决定
D.由负因数的大小决定

C

几个不等于0的有理数相乘，积的符号由负因数的个数决定：当负因数有奇数个时，积为负；当负因数有偶数个时，积为正。

3. 计算：
(1) $(-4)×(+5)×0.25$. (2) $\frac{1}{3}×\frac{1}{2}×(-18)-5×(-\frac{1}{2})$.
(3) $(+2\frac{1}{2})×(-1\frac{1}{2})×(-\frac{4}{5})×(+\frac{2}{3})$.
(4) $(-0.25)×(-\frac{4}{7})×4×(-7)$.

(1) $(-4)×(+5)×0.25$
$=(-4)×0.25×5$
$=(-1)×5$
$=-5$
(2) $\frac{1}{3}×\frac{1}{2}×(-18)-5×(-\frac{1}{2})$
$=\frac{1}{6}×(-18) + \frac{5}{2}$
$=-3 + \frac{5}{2}$
$=-\frac{6}{2} + \frac{5}{2}$
$=-\frac{1}{2}$
(3) $(+2\frac{1}{2})×(-1\frac{1}{2})×(-\frac{4}{5})×(+\frac{2}{3})$
$=(\frac{5}{2})×(-\frac{3}{2})×(-\frac{4}{5})×(\frac{2}{3})$
$=[\frac{5}{2}×(-\frac{4}{5})]×[(-\frac{3}{2})×\frac{2}{3}]$
$=(-2)×(-1)$
$=2$
(4) $(-0.25)×(-\frac{4}{7})×4×(-7)$
$=(-0.25×4)×[(-\frac{4}{7})×(-7)]$
$=(-1)×4$
$=-4$

4. 已知 $a = -3$,$b = -(+4)$,$c = -(-1)$,$d = |-2|$,求 $ab - bc - cd - d$ 的值.

首先计算各个变量的值：
$a = -3$，
$b = -(+4) = -4$，
$c = -(-1) = 1$，
$d = |-2| = 2$，
将$a$，$b$，$c$，$d$的值代入$ab - bc - cd - d$中，
原式$= (-3) × (-4) - (-4) × 1 - 1 × 2 - 2$
$= 12 + 4 - 2 - 2$
$= 12$