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8. 当 $a,b$ 互为相反数时,下列各式可能成立的是
（
A.$\dfrac{b}{a}=-1$
B.$\dfrac{b}{a}=1$
C.$\dfrac{b}{a}=0$
D.$\dfrac{b}{a}≤ 0$
（
A
）A.$\dfrac{b}{a}=-1$
B.$\dfrac{b}{a}=1$
C.$\dfrac{b}{a}=0$
D.$\dfrac{b}{a}≤ 0$
答案
8.A
9. (2024·高新区三模)数轴上表示$a,b$两数的点分别在原点左、右两侧,下列结论一定正确的是（
A.$a+b＞0$
B.$a-b＞0$
C.$a· b＜0$
D.$a÷ b＞0$
C
）A.$a+b＞0$
B.$a-b＞0$
C.$a· b＜0$
D.$a÷ b＞0$
答案
9.C
10. 计算：$(-15)÷(-5)×\dfrac{1}{5}=$
$\dfrac{3}{5}$
.答案
10.$\dfrac{3}{5}$
11. 已知$|x|=4,|y|=\dfrac{1}{2}$，且$xy＜0$，则$\dfrac{x}{y}$的值为
-8
.答案
11.$-8$
12.计算：
(1)$(+1.25)÷(-0.5)÷(-\dfrac{5}{8})$；
(2)$-18÷(+3.25)÷(-2\dfrac{1}{4})$；
(3)$1\dfrac{7}{8}÷(-10)×(-3\dfrac{1}{3})÷(-3\dfrac{3}{4})$；
(4)$-8÷(-\dfrac{1}{2})×\dfrac{5}{16}÷1\dfrac{3}{5}×(-\dfrac{6}{5})$。
(1)$(+1.25)÷(-0.5)÷(-\dfrac{5}{8})$；
(2)$-18÷(+3.25)÷(-2\dfrac{1}{4})$；
(3)$1\dfrac{7}{8}÷(-10)×(-3\dfrac{1}{3})÷(-3\dfrac{3}{4})$；
(4)$-8÷(-\dfrac{1}{2})×\dfrac{5}{16}÷1\dfrac{3}{5}×(-\dfrac{6}{5})$。
答案
12.解:(1)原式=$1.25÷0.5÷\dfrac{5}{8}=\dfrac{5}{4}×2×\dfrac{8}{5}=4$.
(2)原式=$18÷3.25÷2\dfrac{1}{4}=18×\dfrac{4}{13}×\dfrac{4}{9}=\dfrac{32}{13}$.
(3)原式=$-\dfrac{15}{8}×\dfrac{1}{10}×\dfrac{10}{3}×\dfrac{4}{15}=-\dfrac{1}{6}$.
(4)原式=$-8×2×\dfrac{5}{16}×\dfrac{5}{8}×\dfrac{6}{5}=-\dfrac{15}{4}$.
(2)原式=$18÷3.25÷2\dfrac{1}{4}=18×\dfrac{4}{13}×\dfrac{4}{9}=\dfrac{32}{13}$.
(3)原式=$-\dfrac{15}{8}×\dfrac{1}{10}×\dfrac{10}{3}×\dfrac{4}{15}=-\dfrac{1}{6}$.
(4)原式=$-8×2×\dfrac{5}{16}×\dfrac{5}{8}×\dfrac{6}{5}=-\dfrac{15}{4}$.
13. 我们在解题时经常碰到一题多解的情况,如计算: $(-\dfrac{1}{30}) ÷ (\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5})$.
解法一: 原式 $=(-\dfrac{1}{30}) ÷ (\dfrac{5}{6}-\dfrac{1}{2})=-\dfrac{1}{30} × 3=-\dfrac{1}{10}$.
解法二: 因为原式的倒数为 $(\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5}) ÷ (-\dfrac{1}{30})=(\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5}) ×(-30)=\dfrac{2}{3} ×$ $(-30)-\dfrac{1}{10} ×(-30)+\dfrac{1}{6} ×(-30)-\dfrac{2}{5} ×(-30)=-20+3-5+12=-10$,
所以原式 $=-\dfrac{1}{10}$.
根据材料内容,选择合适的方法计算: $(-\dfrac{1}{30}) ÷ (\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{15})$.
解法一: 原式 $=(-\dfrac{1}{30}) ÷ (\dfrac{5}{6}-\dfrac{1}{2})=-\dfrac{1}{30} × 3=-\dfrac{1}{10}$.
解法二: 因为原式的倒数为 $(\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5}) ÷ (-\dfrac{1}{30})=(\dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{6}-\dfrac{2}{5}) ×(-30)=\dfrac{2}{3} ×$ $(-30)-\dfrac{1}{10} ×(-30)+\dfrac{1}{6} ×(-30)-\dfrac{2}{5} ×(-30)=-20+3-5+12=-10$,
所以原式 $=-\dfrac{1}{10}$.
根据材料内容,选择合适的方法计算: $(-\dfrac{1}{30}) ÷ (\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{15})$.
答案
13.解:因为原式的倒数为$(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{15})÷(-\dfrac{1}{30})$
$=(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{15})×(-30)$
$=\dfrac{1}{3}×(-30)+\dfrac{1}{5}×30+\dfrac{1}{15}×30$
$=-10+6+2$
$=-2$.
所以原式$=-\dfrac{1}{2}$.
$=(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{15})×(-30)$
$=\dfrac{1}{3}×(-30)+\dfrac{1}{5}×30+\dfrac{1}{15}×30$
$=-10+6+2$
$=-2$.
所以原式$=-\dfrac{1}{2}$.
