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1. 计算：
(1) $(-2.01)+2.5+1+2.01+(-\dfrac{5}{2})$；
(2) $(-\dfrac{3}{7})×(-\dfrac{4}{5})×(-\dfrac{7}{12})$。
(1) $(-2.01)+2.5+1+2.01+(-\dfrac{5}{2})$；
(2) $(-\dfrac{3}{7})×(-\dfrac{4}{5})×(-\dfrac{7}{12})$。
答案
(1) 原式$=(-2.01)+2.01+2.5+(-\dfrac{5}{2})+1=1$.
(2) 原式$= (-\dfrac{3}{7})× (-\dfrac{7}{12})× (-\dfrac{4}{5}) =\dfrac{1}{4}× (-\dfrac{4}{5}) =-\dfrac{1}{5}$.
(2) 原式$= (-\dfrac{3}{7})× (-\dfrac{7}{12})× (-\dfrac{4}{5}) =\dfrac{1}{4}× (-\dfrac{4}{5}) =-\dfrac{1}{5}$.
2. 计算$(\dfrac{11}{12}-\dfrac{7}{6}+\dfrac{3}{4}-\dfrac{13}{24})×(-24)$的结果是 （
A.1
B.-1
C.10
D.-10
A
）A.1
B.-1
C.10
D.-10
答案
A 【解析】原式$=\frac{11}{12}×(-24)-\frac{7}{6}×(-24)+\frac{3}{4}×(-24)-\frac{13}{24}×(-24)=-22+28-18+13=6-18+13=-12+13=1$.故选A.
3. ★★★ 计算:(1)$17.48×37 - 174.8×1.9 + 1.748×820 = \_\_\_\_\_\_$;
答案
(1)1748 【解析】$17.48×37-174.8×1.9+1.748×820 =17.48×37-17.48×19+17.48×82 =17.48×(37-19+82)=17.48×100=1748$.
(2)$999× 118\dfrac{4}{5}+333× (-\dfrac{3}{5})-999× 18\dfrac{3}{5}=$
99900
.答案
(2)99900 【解析】原式$=999×[ 118\dfrac{4}{5}+(-\dfrac{1}{5})-18\dfrac{3}{5}] =999×100=99900$.
4. （2025·连云港校级月考）阅读下列材料：
计算：$\frac{1}{60}÷( \frac{1}{3}-\frac{1}{4}+\frac{1}{12} )$.
解法①：原式$=\frac{1}{60}÷\frac{1}{3}-\frac{1}{60}÷\frac{1}{4}+\frac{1}{60}÷\frac{1}{12}=\frac{1}{60}×3-\frac{1}{60}×4+\frac{1}{60}×12=\frac{11}{60}$.
解法②：原式的倒数为$( \frac{1}{3}-\frac{1}{4}+\frac{1}{12} )÷\frac{1}{60}=( \frac{1}{3}-\frac{1}{4}+\frac{1}{12} )×60=\frac{1}{3}×60-\frac{1}{4}×60+\frac{1}{12}×60=20-15+5=10$，
所以原式$=\frac{1}{10}$.
（1）上述解法中，你认为解法________是错误的；
（2）计算：$-\frac{1}{42}÷( \frac{3}{7}-\frac{5}{14}+\frac{2}{3}-\frac{1}{6} )$.
$\gg$进一步挑战进阶专题：P20 专题6~P25 专题11
计算：$\frac{1}{60}÷( \frac{1}{3}-\frac{1}{4}+\frac{1}{12} )$.
解法①：原式$=\frac{1}{60}÷\frac{1}{3}-\frac{1}{60}÷\frac{1}{4}+\frac{1}{60}÷\frac{1}{12}=\frac{1}{60}×3-\frac{1}{60}×4+\frac{1}{60}×12=\frac{11}{60}$.
解法②：原式的倒数为$( \frac{1}{3}-\frac{1}{4}+\frac{1}{12} )÷\frac{1}{60}=( \frac{1}{3}-\frac{1}{4}+\frac{1}{12} )×60=\frac{1}{3}×60-\frac{1}{4}×60+\frac{1}{12}×60=20-15+5=10$，
所以原式$=\frac{1}{10}$.
（1）上述解法中，你认为解法________是错误的；
（2）计算：$-\frac{1}{42}÷( \frac{3}{7}-\frac{5}{14}+\frac{2}{3}-\frac{1}{6} )$.
$\gg$进一步挑战进阶专题：P20 专题6~P25 专题11
答案
(1)① 【解析】解法①是错误的,除法没有分配律.
(2)原式的倒数为 $(\dfrac{3}{7}-\dfrac{5}{14}+\dfrac{2}{3}-\dfrac{1}{6})÷ (-\dfrac{1}{42}) = (\dfrac{3}{7}-\dfrac{5}{14}+\dfrac{2}{3}-\dfrac{1}{6})× (-42) = \dfrac{3}{7}× (-42)-\dfrac{5}{14}× (-42)+\dfrac{2}{3}× (-42)-\dfrac{1}{6}× (-42) = -18+15-28+7=-24$,所以原式$=-\dfrac{1}{24}$.
(2)原式的倒数为 $(\dfrac{3}{7}-\dfrac{5}{14}+\dfrac{2}{3}-\dfrac{1}{6})÷ (-\dfrac{1}{42}) = (\dfrac{3}{7}-\dfrac{5}{14}+\dfrac{2}{3}-\dfrac{1}{6})× (-42) = \dfrac{3}{7}× (-42)-\dfrac{5}{14}× (-42)+\dfrac{2}{3}× (-42)-\dfrac{1}{6}× (-42) = -18+15-28+7=-24$,所以原式$=-\dfrac{1}{24}$.
