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1.【教材P45例6变式】计算$2×(-\dfrac{1}{2})÷(-4)$的结果是（
A.1
B.$-1$
C.$\dfrac{1}{4}$
D.$-\dfrac{1}{4}$
C
）A.1
B.$-1$
C.$\dfrac{1}{4}$
D.$-\dfrac{1}{4}$
答案
1.C
2.算式$[(-6.1)-□]÷(-71\frac{1}{2})=0$中，$□$表示的数是（
A.6.1
B.-6.1
C.$71\frac{1}{2}$
D.0
B
）A.6.1
B.-6.1
C.$71\frac{1}{2}$
D.0
答案
2.B 点拨:由题目的特点来看,最后一步运算是[ ]÷( )=0,从而可知[ ]里面结果是0,得$(-6.1)-□=0$,即$□=-6.1$,故选B.
3. 计算$(-\dfrac{1}{8})÷(1\dfrac{1}{2}-\dfrac{3}{2}-\dfrac{1}{4})$的结果是（
A.$\dfrac{1}{2}$
B.$2$
C.$-\dfrac{1}{2}$
D.$-2$
A
）A.$\dfrac{1}{2}$
B.$2$
C.$-\dfrac{1}{2}$
D.$-2$
答案
3.A
4. 下列运算变形正确的是 （
A.$(-13) ÷ (-\dfrac{1}{3}) × (-3) = 13 × [\dfrac{1}{3} × (-3)]$
B.$(-13) ÷ (-\dfrac{1}{3}) × (-3) = -13 × 3 × 3$
C.$(-13) ÷ (-\dfrac{1}{3}) × (-3) = 13 ÷ \dfrac{1}{3} × 3$
D.$(-13) ÷ (-\dfrac{1}{3}) × (-3) = -13 × \dfrac{1}{3} × 3$
B
）A.$(-13) ÷ (-\dfrac{1}{3}) × (-3) = 13 × [\dfrac{1}{3} × (-3)]$
B.$(-13) ÷ (-\dfrac{1}{3}) × (-3) = -13 × 3 × 3$
C.$(-13) ÷ (-\dfrac{1}{3}) × (-3) = 13 ÷ \dfrac{1}{3} × 3$
D.$(-13) ÷ (-\dfrac{1}{3}) × (-3) = -13 × \dfrac{1}{3} × 3$
答案
4.B
5.【教材P47练习3变式】用计算器计算下列各题：
(1)－98×(－32.7)÷0.6；
(2)－0.386×1.63＋2.68÷0.2.(结果保留两位小数)
(1)－98×(－32.7)÷0.6；
(2)－0.386×1.63＋2.68÷0.2.(结果保留两位小数)
答案
5.(1)5341 (2)12.77
1. 计算$\frac{1}{32} ÷ ( -\frac{1}{8} + \frac{5}{4} - \frac{3}{2} + \frac{1}{16} )$的结果是 （
A.$-10$
B.$-\frac{1}{10}$
C.$10$
D.$\frac{1}{10}$
B
）A.$-10$
B.$-\frac{1}{10}$
C.$10$
D.$\frac{1}{10}$
答案
1.B 点拨:因为$(-\frac{1}{8}+\frac{5}{4}-\frac{3}{2}+\frac{1}{16})÷\frac{1}{32}=(-\frac{1}{8}+\frac{5}{4}-\frac{3}{2}+\frac{1}{16})×32=-\frac{1}{8}×32+\frac{5}{4}×32-\frac{3}{2}×32+\frac{1}{16}×32=-4+40-48+2=-10$.所以原式$=-\frac{1}{10}$.故选B.
