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10.计算.
(1)$(-5)-(-5)×\frac{1}{10}÷\frac{1}{10}×(-5)$；
(2)$-1^4-(1-\frac{1}{3})÷3×[2-(-3)^2]$.
(1)$(-5)-(-5)×\frac{1}{10}÷\frac{1}{10}×(-5)$；
(2)$-1^4-(1-\frac{1}{3})÷3×[2-(-3)^2]$.
答案
10.【解】(1)原式$=(-5)+\frac{1}{2}×10×(-5)$
$=(-5)+(-25)$
$=-30$.
(2)原式$=-1-\frac{2}{3}÷3×(2-9)$
$=-1-\frac{2}{3}×\frac{1}{3}×(-7)$
$=-1+\frac{14}{9}$
$=\frac{5}{9}$.
$=(-5)+(-25)$
$=-30$.
(2)原式$=-1-\frac{2}{3}÷3×(2-9)$
$=-1-\frac{2}{3}×\frac{1}{3}×(-7)$
$=-1+\frac{14}{9}$
$=\frac{5}{9}$.
11. 运用简便方法计算：
(1) $2\dfrac{1}{7} - 3\dfrac{2}{3} -5\dfrac{1}{3} + (-3\dfrac{1}{7})$；
(2) $(\dfrac{1}{4})^2 ÷ (-2\dfrac{1}{2}) + (11\dfrac{1}{4} + 2\dfrac{1}{3} -13\dfrac{3}{4}) ×24 - \dfrac{1}{(-0.2)^3}$。
(1) $2\dfrac{1}{7} - 3\dfrac{2}{3} -5\dfrac{1}{3} + (-3\dfrac{1}{7})$；
(2) $(\dfrac{1}{4})^2 ÷ (-2\dfrac{1}{2}) + (11\dfrac{1}{4} + 2\dfrac{1}{3} -13\dfrac{3}{4}) ×24 - \dfrac{1}{(-0.2)^3}$。
答案
11.【解】(1)原式$=2\dfrac{1}{7}-3\dfrac{2}{3}-5\dfrac{1}{3}-3\dfrac{1}{7}$
$=(2\dfrac{1}{7}-3\dfrac{1}{7})+(-3\dfrac{2}{3}-5\dfrac{1}{3})$
$=-1-9$
$=-10$.
(2)原式$=\dfrac{1}{16} × (-\dfrac{2}{5}) + (\dfrac{45}{4} + \dfrac{7}{3} - \dfrac{55}{4}) × 24-\dfrac{1}{(-\dfrac{1}{5})^3}$
$=-\dfrac{1}{40}+\dfrac{45}{4}×24+\dfrac{7}{3}×24-\dfrac{55}{4}×24+125$
$=-\dfrac{1}{40}+(270+56-330+125)$
$=-\dfrac{1}{40}+121$
$=120\dfrac{39}{40}$.
$=(2\dfrac{1}{7}-3\dfrac{1}{7})+(-3\dfrac{2}{3}-5\dfrac{1}{3})$
$=-1-9$
$=-10$.
(2)原式$=\dfrac{1}{16} × (-\dfrac{2}{5}) + (\dfrac{45}{4} + \dfrac{7}{3} - \dfrac{55}{4}) × 24-\dfrac{1}{(-\dfrac{1}{5})^3}$
$=-\dfrac{1}{40}+\dfrac{45}{4}×24+\dfrac{7}{3}×24-\dfrac{55}{4}×24+125$
$=-\dfrac{1}{40}+(270+56-330+125)$
$=-\dfrac{1}{40}+121$
$=120\dfrac{39}{40}$.
12. 用简便方法计算：$(-3) × (-\frac{1}{4}) + 0.25 × 24.5 + (-5\frac{1}{2}) × (-25\%)$
答案
12.【解】原式$=3×\dfrac{1}{4}+\dfrac{1}{4}×\dfrac{49}{2}+\dfrac{11}{2}×\dfrac{1}{4}$
$=(3+\dfrac{49}{2}+\dfrac{11}{2})×\dfrac{1}{4}$
$=33×\dfrac{1}{4}$
$=\dfrac{33}{4}$.
$=(3+\dfrac{49}{2}+\dfrac{11}{2})×\dfrac{1}{4}$
$=33×\dfrac{1}{4}$
$=\dfrac{33}{4}$.
13. 阅读以下材料，完成相关的填空和计算.
(1)根据倒数的定义我们知道，若$(a+b)÷c=-2$，则$c÷(a+b)=\_\_\_\_\_\_$；
(2)计算：$(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})÷\dfrac{1}{36}$；
(3)根据以上信息可知，$(-\dfrac{1}{36})÷(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})=\_\_\_\_\_\_$.
(1)根据倒数的定义我们知道，若$(a+b)÷c=-2$，则$c÷(a+b)=\_\_\_\_\_\_$；
(2)计算：$(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})÷\dfrac{1}{36}$；
(3)根据以上信息可知，$(-\dfrac{1}{36})÷(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})=\_\_\_\_\_\_$.
答案
13.【解】(1)$-\dfrac{1}{2}$
(2)$(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})÷\dfrac{1}{36}$
$=(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})×36$
$=\dfrac{5}{12}×36-\dfrac{1}{9}×36+\dfrac{2}{3}×36$
$=15-4+24$
$=35$.
(3)$-\dfrac{1}{35}$
【点拨】由(2)得$(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})÷\dfrac{1}{36}=35$,
所以$\dfrac{1}{36}÷(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})=\dfrac{1}{35}$.
所以$(-\dfrac{1}{36})÷(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})=-\dfrac{1}{35}$.
(2)$(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})÷\dfrac{1}{36}$
$=(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})×36$
$=\dfrac{5}{12}×36-\dfrac{1}{9}×36+\dfrac{2}{3}×36$
$=15-4+24$
$=35$.
(3)$-\dfrac{1}{35}$
【点拨】由(2)得$(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})÷\dfrac{1}{36}=35$,
所以$\dfrac{1}{36}÷(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})=\dfrac{1}{35}$.
所以$(-\dfrac{1}{36})÷(\dfrac{5}{12}-\dfrac{1}{9}+\dfrac{2}{3})=-\dfrac{1}{35}$.
14. 计算:$89+899+8\ 999+89\ 999-9-99-999-9\ 999-99\ 999.$
答案
14.【解】原式$=(90+900+9\ 000+90\ 000-4)-(10+100+1\ 000+10\ 000+100\ 000-5)=$
$99\ 990-111\ 110-4+5=-11\ 119$.
$99\ 990-111\ 110-4+5=-11\ 119$.
15. 计算:$1-3-5+7+9-11-13+15+17-\dots-2\ 021+2\ 023+2\ 025-2\ 027-2\ 029+2\ 031.$
答案
15.【解】原式$=(1-3-5+7)+(9-11-13+15)+\dots+(2\ 017-2\ 019-2\ 021+2\ 023)+$
$(2\ 025-2\ 027-2\ 029+2\ 031)=0$.
$(2\ 025-2\ 027-2\ 029+2\ 031)=0$.
