9.用简便方法计算：$-19\frac{15}{16}×8$
答案：9.解:原式$=-(20-\dfrac{1}{16})×8=-(20×8-\dfrac{1}{16}×8)=-(160-\dfrac{1}{2})=-159\dfrac{1}{2}.$
10. 计算：
(1)$(-12.5)×(-2.5)×0.5×(-8)×(-4)×2$；
(2)$(-\dfrac{2}{13})×(-\dfrac{7}{8})×\dfrac{26}{49}×\dfrac{8}{7}×(-3)$；
(3)$(-15.3)×(-6.19)+(-15.3)×16.19-153×(-2)$；
(4)$20\dfrac{1}{18}×(-9)$；
(5)$(\dfrac{1}{2}-1-\dfrac{3}{4}+\dfrac{5}{6}-\dfrac{7}{12})×(-12)$；
(6)$(-3\dfrac{1}{7})×(3\dfrac{1}{7}-7\dfrac{1}{3})×(-\dfrac{7}{22})×\dfrac{21}{22}$。
答案：10.解:(1)原式$=12.5×2.5×0.5×8×4×2=(12.5×8)×(2.5×4)×(0.5×2)=100×10×1=1000.$
(2)原式$=-\dfrac{2}{13}×\dfrac{7}{8}×\dfrac{26}{49}×\dfrac{8}{7}×3=-(\dfrac{2}{13}×\dfrac{26}{49})×(\dfrac{7}{8}×\dfrac{8}{7})×3=-\dfrac{4}{49}×1×3=-\dfrac{12}{49}.$
(3)原式$=15.3×6.19-15.3×16.19+15.3×20=15.3×(6.19-16.19+20)=15.3×10=153.$
(4)原式$=-(20+\dfrac{1}{18})×9=-(20×9+\dfrac{1}{18}×9)=-(180+\dfrac{1}{2})=-180\dfrac{1}{2}.$
(5)原式$=\dfrac{1}{2}×(-12)-1×(-12)-\dfrac{3}{4}×(-12)+\dfrac{5}{6}×(-12)-\dfrac{7}{12}×(-12)=-6+12+9-10+7=12.$
(6)原式$=[(-\dfrac{22}{7})×(-\dfrac{7}{22})]×(\dfrac{22}{7}-\dfrac{22}{3})×\dfrac{21}{22}=(\dfrac{22}{7}-\dfrac{22}{3})×\dfrac{21}{22}=\dfrac{22}{7}×\dfrac{21}{22}-\dfrac{22}{3}×\dfrac{21}{22}=3-7=-4.$
11.已知a,b互为相反数，c的倒数是4，d的绝对值是最小的正整数.
求:(1)$3a+3b-4c$的值；
(2)$8c-d+cd$的值.
答案：11.解:由题意,得$a+b=0,c=\dfrac{1}{4},|d|=1$,所以$d=±1.$
(1)$3a+3b-4c=3(a+b)-4c=3×0-4×\dfrac{1}{4}=0-1=-1.$
(2)当$c=\dfrac{1}{4},d=1$时,
$8c-d+cd=8×\dfrac{1}{4}-1+\dfrac{1}{4}×1=2-1+\dfrac{1}{4}=\dfrac{5}{4};$
当$c=\dfrac{1}{4},d=-1$时,
$8c-d+cd=8×\dfrac{1}{4}-(-1)+\dfrac{1}{4}×(-1)=2+1-\dfrac{1}{4}=\dfrac{11}{4}.$
综上所述,$8c-d+cd$的值为$\dfrac{5}{4}$或$\dfrac{11}{4}.$
