1. 乘法交换律:两个有理数相乘,交换乘数的位置,积
2. 乘法结合律:三个有理数相乘,先把前两个数相乘,或者先把后两个数相乘,积
3. 分配律:一个数与两个数的和相乘,等于把这个数分别与这两个数相乘,再把积
不变
,即$ab = ba$.2. 乘法结合律:三个有理数相乘,先把前两个数相乘,或者先把后两个数相乘,积
不变
,即$(ab)c = a(bc)$.3. 分配律:一个数与两个数的和相乘,等于把这个数分别与这两个数相乘,再把积
相加
,即$a(b + c) = $$ab + ac$
.答案：1. 不变 2. 不变 3. 相加 $ ab + ac $
1. 在算式$1.25×(-\frac {3}{4})×(-8)= 1.25×(-8)×(-\frac {3}{4})= [1.25×(-8)]×(-\frac {3}{4})$中,运用了 (
A.分配律
B.分配律和乘法结合律
C.乘法交换律和乘法结合律
D.乘法交换律和分配律
C
)A.分配律
B.分配律和乘法结合律
C.乘法交换律和乘法结合律
D.乘法交换律和分配律
答案：C
解析：
解：算式中，将$-\frac{3}{4}$与$-8$交换位置，运用了乘法交换律；再将$1.25$与$-8$结合相乘，运用了乘法结合律。
答案：C
答案：C
2. 填空:
(1)$10×(-8)=$
(2)$[12×(-2)]×(-5)= 12×[$
(3)$(-24)×(\frac {1}{6}-\frac {1}{8})= (-24)×$
(1)$10×(-8)=$
$(-8)$
$×10$;(2)$[12×(-2)]×(-5)= 12×[$
$(-2)$
$×(-5)]$;(3)$(-24)×(\frac {1}{6}-\frac {1}{8})= (-24)×$
$\frac{1}{6}$
$+(-24)×$$\left(-\frac{1}{8}\right)$
.答案：(1) $(-8)$ (2) $(-2)$ (3) $\frac{1}{6}$ $\left(-\frac{1}{8}\right)$
3. 计算:$(-13)×(-1+\frac {4}{13})=$
9
.答案：9
解析：
$(-13)×(-1+\frac {4}{13})$
$=(-13)×(-1)+(-13)×\frac {4}{13}$
$=13 - 4$
$=9$
$=(-13)×(-1)+(-13)×\frac {4}{13}$
$=13 - 4$
$=9$
4. 计算:
(1)$27.9×(-5)×(-2)$;
(2)$-3×(-\frac {7}{5})×(-\frac {2}{3})×\frac {4}{7}$;
(3)$1.25×(-\frac {71}{20})×(-8)$;
(4)$(\frac {1}{3}+\frac {1}{9}-\frac {5}{12})×(-108)$.
(1)$27.9×(-5)×(-2)$;
(2)$-3×(-\frac {7}{5})×(-\frac {2}{3})×\frac {4}{7}$;
(3)$1.25×(-\frac {71}{20})×(-8)$;
(4)$(\frac {1}{3}+\frac {1}{9}-\frac {5}{12})×(-108)$.
答案：(1)解：$27.9×(-5)×(-2)$
$=27.9×[(-5)×(-2)]$
$=27.9×10$
$=279$
(2)解：$-3×(-\frac {7}{5})×(-\frac {2}{3})×\frac {4}{7}$
$=[-3×(-\frac {2}{3})]×[(-\frac {7}{5})×\frac {4}{7}]$
$=2×(-\frac {4}{5})$
$=-\frac {8}{5}$
(3)解：$1.25×(-\frac {71}{20})×(-8)$
$=1.25×(-8)×(-\frac {71}{20})$
$=(-10)×(-\frac {71}{20})$
$=\frac {71}{2}$
(4)解：$(\frac {1}{3}+\frac {1}{9}-\frac {5}{12})×(-108)$
$=\frac {1}{3}×(-108)+\frac {1}{9}×(-108)-\frac {5}{12}×(-108)$
$=-36-12+45$
$=-3$
$=27.9×[(-5)×(-2)]$
$=27.9×10$
$=279$
(2)解：$-3×(-\frac {7}{5})×(-\frac {2}{3})×\frac {4}{7}$
$=[-3×(-\frac {2}{3})]×[(-\frac {7}{5})×\frac {4}{7}]$
$=2×(-\frac {4}{5})$
$=-\frac {8}{5}$
(3)解：$1.25×(-\frac {71}{20})×(-8)$
$=1.25×(-8)×(-\frac {71}{20})$
$=(-10)×(-\frac {71}{20})$
$=\frac {71}{2}$
(4)解：$(\frac {1}{3}+\frac {1}{9}-\frac {5}{12})×(-108)$
$=\frac {1}{3}×(-108)+\frac {1}{9}×(-108)-\frac {5}{12}×(-108)$
$=-36-12+45$
$=-3$
解析：
(1)解：$27.9×(-5)×(-2)$
$=27.9×[(-5)×(-2)]$
$=27.9×10$
$=279$
(2)解：$-3×(-\frac {7}{5})×(-\frac {2}{3})×\frac {4}{7}$
$=[-3×(-\frac {2}{3})]×[(-\frac {7}{5})×\frac {4}{7}]$
$=2×(-\frac {4}{5})$
$=-\frac {8}{5}$
(3)解：$1.25×(-\frac {71}{20})×(-8)$
$=1.25×(-8)×(-\frac {71}{20})$
$=(-10)×(-\frac {71}{20})$
$=\frac {71}{2}$
(4)解：$(\frac {1}{3}+\frac {1}{9}-\frac {5}{12})×(-108)$
$=\frac {1}{3}×(-108)+\frac {1}{9}×(-108)-\frac {5}{12}×(-108)$
$=-36-12+45$
$=-3$
$=27.9×[(-5)×(-2)]$
$=27.9×10$
$=279$
(2)解：$-3×(-\frac {7}{5})×(-\frac {2}{3})×\frac {4}{7}$
$=[-3×(-\frac {2}{3})]×[(-\frac {7}{5})×\frac {4}{7}]$
$=2×(-\frac {4}{5})$
$=-\frac {8}{5}$
(3)解：$1.25×(-\frac {71}{20})×(-8)$
$=1.25×(-8)×(-\frac {71}{20})$
$=(-10)×(-\frac {71}{20})$
$=\frac {71}{2}$
(4)解：$(\frac {1}{3}+\frac {1}{9}-\frac {5}{12})×(-108)$
$=\frac {1}{3}×(-108)+\frac {1}{9}×(-108)-\frac {5}{12}×(-108)$
$=-36-12+45$
$=-3$