10.已知有理数$a=(-2)×3$,则$a$的倒数是(
A.$-\frac{1}{6}$
B.$\frac{1}{6}$
C.$-6$
D.$6$
A
)A.$-\frac{1}{6}$
B.$\frac{1}{6}$
C.$-6$
D.$6$
答案
A
11.(2024年北京市)有理数a、b在数轴上的对应点的位置如图所示,下列结论正确的是
(

A.$b>-1$
B.$|b|>2$
C.$a+b>0$
D.$ab>0$
(
C
)A.$b>-1$
B.$|b|>2$
C.$a+b>0$
D.$ab>0$
答案
C
12.若“!”是一种数学运算符号,并且1!=1,2!=2×1=2,3!=3×2×1=6,4!=4×3×2×1,…,则$\frac{100!}{98!}$的值为
9900
.答案
9900
13.计算:
(1)$(-30)×(-\dfrac{1}{6})×(+0.2)×6$;
(2)$\dfrac{1}{2}×(-\dfrac{2}{3})×(-3\dfrac{1}{4})×(-1\dfrac{1}{3})$;
(3)$\dfrac{1}{4}×(-16)×(-\dfrac{4}{5})×\left|-1\dfrac{1}{4}\right|$。
(1)$(-30)×(-\dfrac{1}{6})×(+0.2)×6$;
(2)$\dfrac{1}{2}×(-\dfrac{2}{3})×(-3\dfrac{1}{4})×(-1\dfrac{1}{3})$;
(3)$\dfrac{1}{4}×(-16)×(-\dfrac{4}{5})×\left|-1\dfrac{1}{4}\right|$。
答案
(1) 解:原式=$30×\dfrac{1}{6}×\dfrac{1}{5}×6=6$.
(2) 解:原式=$-(\dfrac{1}{2}×\dfrac{2}{3}×\dfrac{13}{4}×\dfrac{4}{3})=-\dfrac{13}{9}$.
(3) 解:原式=$\dfrac{1}{4}×16×\dfrac{4}{5}×\dfrac{5}{4}=4$.
(2) 解:原式=$-(\dfrac{1}{2}×\dfrac{2}{3}×\dfrac{13}{4}×\dfrac{4}{3})=-\dfrac{13}{9}$.
(3) 解:原式=$\dfrac{1}{4}×16×\dfrac{4}{5}×\dfrac{5}{4}=4$.
14.若定义一种新的运算“*”,规定有理数$a*b=4ab$,如$2*3=4×2×3=24$。
(1)求$3*(-4)$的值;
(2)求$(-2)*(6*3)$的值。
(1)求$3*(-4)$的值;
(2)求$(-2)*(6*3)$的值。
答案
(1) 解:$3 * (-4)=4×3×(-4)=-48$.
(2) $(-2) * (6 * 3)=(-2) * (4×6×3)=(-2) * 72=4×(-2)×72=-576$.
(2) $(-2) * (6 * 3)=(-2) * (4×6×3)=(-2) * 72=4×(-2)×72=-576$.
15.一辆汽车沿一条东西走向的公路行驶,它从A地沿这条公路向东以40km/h的速度行驶了2.5h,又反向以45km/h的速度行驶了2h,到达B地.问B地在A地的东边还是西边?它们之间的距离是多少千米?(规定向东为正,向西为负)
答案
解:$40×2.5+(-45×2)=10(\mathrm{km})$.答:B地在A地的东边,相距10km.
16.(核心素养·应用意识)观察下列各式:
$\frac{1}{2}×\frac{2}{3}=\frac{1}{3},$
$\frac{1}{2}×\frac{2}{3}×\frac{3}{4}=\frac{1}{4},$
$\frac{1}{2}×\frac{2}{3}×\frac{3}{4}×\frac{4}{5}=\frac{1}{5},$
……
(1)猜想$\frac{1}{2}×\frac{2}{3}×\frac{3}{4}×…×\frac{49}{50}=$
(2)根据以上规律,计算:$(\frac{1}{100}-1)×(\frac{1}{99}-1)×(\frac{1}{98}-1)×…×(\frac{1}{2}-1).$
$\frac{1}{2}×\frac{2}{3}=\frac{1}{3},$
$\frac{1}{2}×\frac{2}{3}×\frac{3}{4}=\frac{1}{4},$
$\frac{1}{2}×\frac{2}{3}×\frac{3}{4}×\frac{4}{5}=\frac{1}{5},$
……
(1)猜想$\frac{1}{2}×\frac{2}{3}×\frac{3}{4}×…×\frac{49}{50}=$
$\dfrac{1}{50}$
;(2)根据以上规律,计算:$(\frac{1}{100}-1)×(\frac{1}{99}-1)×(\frac{1}{98}-1)×…×(\frac{1}{2}-1).$
答案
(1) $\dfrac{1}{50}$
(2) 解:原式=$-\dfrac{99}{100}×(-\dfrac{98}{99})×(-\dfrac{97}{98})×…×(-\dfrac{1}{2})=-\dfrac{1}{100}$.
(2) 解:原式=$-\dfrac{99}{100}×(-\dfrac{98}{99})×(-\dfrac{97}{98})×…×(-\dfrac{1}{2})=-\dfrac{1}{100}$.
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