登录
1. (2026·信阳期中)把$(-2)-(+3)-(-6)+(-5)$统一为加法运算，正确的是（
A.$(-2)+(+3)+(-6)+(-5)$
B.$(-2)+(-3)+(+6)+(-5)$
C.$(-2)+(+3)+(+6)+(+5)$
D.$(-2)+(-3)+(-6)+(+5)$
B
）A.$(-2)+(+3)+(-6)+(-5)$
B.$(-2)+(-3)+(+6)+(-5)$
C.$(-2)+(+3)+(+6)+(+5)$
D.$(-2)+(-3)+(-6)+(+5)$
答案
1. B 解析:$(-2)-(+3)-(-6)+(-5)=(-2)+(-3)+(+6)+(-5)$,故选 B.
2. 某天早晨气温是$-3\ °\mathrm{C}$,到中午升高了$5\ °\mathrm{C}$,晚上又降低了$3\ °\mathrm{C}$,到午夜又降低了$4\ °\mathrm{C}$,午夜时温度为 (
A.$5\ °\mathrm{C}$
B.$15\ °\mathrm{C}$
C.$-5\ °\mathrm{C}$
D.$1\ °\mathrm{C}$
C
)A.$5\ °\mathrm{C}$
B.$15\ °\mathrm{C}$
C.$-5\ °\mathrm{C}$
D.$1\ °\mathrm{C}$
答案
2. C 解析:根据题意得$-3+5-3-4=-(3+3+4)+5=-10+5=-5(℃)$,则午夜时温度为$-5\ ℃$,故选 C.
3. 教材P39例6变式 计算：
(1) $(-7)-(+5)+(+13)=$
(2) $(-1.6)+(-2.4)-(-7.7)=$
(3) $(-35)+(-22)-(-35)-8=$
(4) $\frac{1}{2}+(-\frac{2}{3})+\frac{4}{5}+(-\frac{1}{2})+(-\frac{1}{3})=$
(1) $(-7)-(+5)+(+13)=$
1
；(2) $(-1.6)+(-2.4)-(-7.7)=$
3.7
；(3) $(-35)+(-22)-(-35)-8=$
-30
；(4) $\frac{1}{2}+(-\frac{2}{3})+\frac{4}{5}+(-\frac{1}{2})+(-\frac{1}{3})=$
$-\frac{1}{5}$
。答案
3. (1)1 解析:原式$=-7-5+13=-12+13=1$.
(2)3.7 解析:原式$=-1.6-2.4+7.7=-4+7.7=3.7$.
(3)-30 解析:原式$=(-35+35)+(-22-8)=0-30=-30$.
(4)$-\frac{1}{5}$ 解析:原式$= (\frac{1}{2}-\frac{1}{2})+(-\frac{2}{3}-\frac{1}{3})+\frac{4}{5}=0-1+\frac{4}{5}=-\frac{1}{5}$.
(2)3.7 解析:原式$=-1.6-2.4+7.7=-4+7.7=3.7$.
(3)-30 解析:原式$=(-35+35)+(-22-8)=0-30=-30$.
(4)$-\frac{1}{5}$ 解析:原式$= (\frac{1}{2}-\frac{1}{2})+(-\frac{2}{3}-\frac{1}{3})+\frac{4}{5}=0-1+\frac{4}{5}=-\frac{1}{5}$.
4. 某公交车上原有22人,经过3个站点时上、下车情况如下(上车记为正,下车记为负):
(+3,-7),(+6,-4),(+2,-1),则现在车上还有
(+3,-7),(+6,-4),(+2,-1),则现在车上还有
21
人.答案
4. 21 解析:由题意得 $22+3-7+6-4+2-1=21$(人).
5. 计算:
(1) $(-7)-(-10)+(-8)-(+2)$;
(2) $(-1.5)-(-0.75)+(+0.25)-(+2.5)$;
(3) $(+3.4)+(-2\dfrac{7}{8})-(-5\dfrac{3}{5})-(+0.125)$;
(4) $(-11\dfrac{2}{3})-(-7\dfrac{2}{5})-\left|-12\dfrac{1}{3}\right|-(-4.6)$.
(1) $(-7)-(-10)+(-8)-(+2)$;
(2) $(-1.5)-(-0.75)+(+0.25)-(+2.5)$;
(3) $(+3.4)+(-2\dfrac{7}{8})-(-5\dfrac{3}{5})-(+0.125)$;
(4) $(-11\dfrac{2}{3})-(-7\dfrac{2}{5})-\left|-12\dfrac{1}{3}\right|-(-4.6)$.
答案
5. (1)原式$=-7+10-8-2=-7$.
(2)原式$=-1.5+0.75+0.25-2.5=-(1.5+2.5)+(0.75+0.25)=-4+1=-3$.
(3)原式$= (+3\dfrac{2}{5})-(-5\dfrac{3}{5})+(-2\dfrac{7}{8})-(+\dfrac{1}{8}) = (+3\dfrac{2}{5})+5\dfrac{3}{5}+ [ (-2\dfrac{7}{8})-(+\dfrac{1}{8}) ] =9-3=6$.
(4)原式$=-11\dfrac{2}{3}+7\dfrac{2}{5}-12\dfrac{1}{3}+4\dfrac{3}{5} = (-11\dfrac{2}{3}-12\dfrac{1}{3})+(7\dfrac{2}{5}+4\dfrac{3}{5}) =-24+12=-12$.
(2)原式$=-1.5+0.75+0.25-2.5=-(1.5+2.5)+(0.75+0.25)=-4+1=-3$.
(3)原式$= (+3\dfrac{2}{5})-(-5\dfrac{3}{5})+(-2\dfrac{7}{8})-(+\dfrac{1}{8}) = (+3\dfrac{2}{5})+5\dfrac{3}{5}+ [ (-2\dfrac{7}{8})-(+\dfrac{1}{8}) ] =9-3=6$.
(4)原式$=-11\dfrac{2}{3}+7\dfrac{2}{5}-12\dfrac{1}{3}+4\dfrac{3}{5} = (-11\dfrac{2}{3}-12\dfrac{1}{3})+(7\dfrac{2}{5}+4\dfrac{3}{5}) =-24+12=-12$.
6. 有理数-7,-3,+5 的和比它们的绝对值的和小（
A.2
B.7
C.15
D.20
D
）A.2
B.7
C.15
D.20
答案
6. D 解析:$(|-7|+|-3|+|+5|)-(-7-3+5)=15-(-5)=20$.
7. 新题型 新运算（2025·南阳期末）大家都知道，七点五十可以说成差十分钟八点，有时这样表达更清楚，这也启发人们设计了一种新的加减记数法。比如：8写成$1\overline{2}$，$1\overline{2}=10-2$；189写成$2\overline{2}9=200-20+9$；7683写成$1\overline{232}3=10000-2320+3$，按这个方法请计算$52\overline{31}-3\overline{24}1=$（
A.2 408
B.1 990
C.2 410
D.3 024
A
）A.2 408
B.1 990
C.2 410
D.3 024
答案
7. A 解析:原式$=(5\ 200-31)-(3\ 000-240+1)=2\ 408$.故选 A.
8. 规定一种运算:如果抽到圆形卡片就加上它上面的数,如果抽到正方形卡片就减去它上面的数,小林抽到的四张卡片如图,则运算结果是
$-\frac{5}{4}$
.答案
8. $-\frac{5}{4}$ 解析:$-(-\frac{1}{4})+(-\frac{5}{8})+(-\frac{1}{8})-\frac{3}{4}=\frac{1}{4}-\frac{5}{8}-\frac{1}{8}-\frac{3}{4}=(\frac{1}{4}-\frac{3}{4})-(\frac{5}{8}+\frac{1}{8})=-\frac{1}{2}-\frac{3}{4}=-\frac{5}{4}$.
