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1.「2026河南郑州期中」将$5-(-3)-(+7)+(-2)$中的减法改成加法，并写成省略加号和括号的形式是（
A.$-5-3+7+2$
B.$5-3-7-2$
C.$5+3-7+2$
D.$5+3-7-2$
D
）A.$-5-3+7+2$
B.$5-3-7-2$
C.$5+3-7+2$
D.$5+3-7-2$
答案
1.D $5-(-3)-(+7)+(-2)=5+3-7-2$.故选 D.
师大版
2.对式子“-8+16-3-6”的读法正确的是 （
A.负8加16减3减6
B.负8正16负3减6
C.负8、加16、负3、负6的和
D.减8加16减3减6
2.对式子“-8+16-3-6”的读法正确的是 （
A
）A.负8加16减3减6
B.负8正16负3减6
C.负8、加16、负3、负6的和
D.减8加16减3减6
答案
2.A 式子“$-8+16-3-6$”可以读作负8、正16、负3、负6的和,还可以读作负8加16减3减6.故选 A.
3.式子-2-1+6-9可以读作
负2、负1、正6、负9的和
，也可以读作负2减1加6减9
。答案
3.答案 负2、负1、正6、负9的和；负2减1加6减9
4.「2026 河北廊坊期末」请指出下面计算从哪一步开始出错（
$1+\dfrac{4}{5}-(+\dfrac{2}{3})-(-\dfrac{1}{5})-(+1\dfrac{1}{3})$
$=1\dfrac{4}{5}-\dfrac{2}{3}+\dfrac{1}{5}-1\dfrac{1}{3}·······································$ ①
$=(1\dfrac{4}{5}+\dfrac{1}{5})-(\dfrac{2}{3}-1\dfrac{1}{3})·······································$ ②
$=2-(-\dfrac{2}{3})·······································$ ③
$=2\dfrac{2}{3}·······································$ ④
A.①
B.②
C.③
D.④
B
）$1+\dfrac{4}{5}-(+\dfrac{2}{3})-(-\dfrac{1}{5})-(+1\dfrac{1}{3})$
$=1\dfrac{4}{5}-\dfrac{2}{3}+\dfrac{1}{5}-1\dfrac{1}{3}·······································$ ①
$=(1\dfrac{4}{5}+\dfrac{1}{5})-(\dfrac{2}{3}-1\dfrac{1}{3})·······································$ ②
$=2-(-\dfrac{2}{3})·······································$ ③
$=2\dfrac{2}{3}·······································$ ④
A.①
B.②
C.③
D.④
答案
4.B 由计算过程可得,从第②步开始出错,应为$(1 \dfrac{4}{5}+\dfrac{1}{5})+(-\dfrac{2}{3}-1 \dfrac{1}{3})$.故选 B.
5.「2026山东滨州期末」若“方框”
表示运算$x - y + z + w$，则“方框”
=
表示运算$x - y + z + w$，则“方框”=-8
。答案
5.答案 -8
解析 由题意得，$\begin{vmatrix} -2&3\\ 3&-6 \end{vmatrix}=-2-3+(-6)+3=-8.$
解析 由题意得，$\begin{vmatrix} -2&3\\ 3&-6 \end{vmatrix}=-2-3+(-6)+3=-8.$
6.计算:
(1)$(-20)+(+12)-(-5)-(+7).$
(2)$3\frac{1}{2}+(-\frac{1}{2})-(-\frac{1}{3})+2\frac{2}{3}.$
(3)$5.6-0.9+4.4+(-8.1)-(+0.1).$
(4)$(+3.5)-(+\frac{2}{5})-(-6.5)+(-\frac{3}{5}).$
(5)$(-3\frac{2}{3})-(-2\frac{3}{4})-(-1\frac{2}{3})-(+1.75).$
(6)$\frac{2}{5}-\left|-1\frac{1}{2}\right|-(+2\frac{1}{4})-(-2.75).$
(1)$(-20)+(+12)-(-5)-(+7).$
(2)$3\frac{1}{2}+(-\frac{1}{2})-(-\frac{1}{3})+2\frac{2}{3}.$
(3)$5.6-0.9+4.4+(-8.1)-(+0.1).$
(4)$(+3.5)-(+\frac{2}{5})-(-6.5)+(-\frac{3}{5}).$
(5)$(-3\frac{2}{3})-(-2\frac{3}{4})-(-1\frac{2}{3})-(+1.75).$
(6)$\frac{2}{5}-\left|-1\frac{1}{2}\right|-(+2\frac{1}{4})-(-2.75).$
答案
6.解析 (1)$(-20)+(+12)-(-5)-(+7)$
$=(-20)+(+12)+(+5)+(-7)=-10.$
(2)$3 \frac{1}{2}+(-\frac{1}{2})-(-\frac{1}{3})+2 \frac{2}{3}$
$=(3 \frac{1}{2}-\frac{1}{2})+(\frac{1}{3}+2 \frac{2}{3})=3+3=6.$
(3)$5.6-0.9+4.4+(-8.1)-(+0.1)$
$=(5.6+4.4)+(-0.9-8.1-0.1)$
$=10-9.1=0.9.$
(4)$(+3.5)-(+\frac{2}{5})-(-6.5)+(-\frac{3}{5})$
$=3.5-\frac{2}{5}+6.5-\frac{3}{5}$
$=3.5+6.5-\frac{2}{5}-\frac{3}{5}$
$=10-1=9.$
(5)$(-3 \frac{2}{3})-(-2 \frac{3}{4})-(-1 \frac{2}{3})-(+1.75)$
$=-3 \frac{2}{3}+2 \frac{3}{4}+1 \frac{2}{3}+(-1 \frac{3}{4})$
$=(-3 \frac{2}{3}+1 \frac{2}{3})+[(+2 \frac{3}{4})+(-1 \frac{3}{4})]$
$=-2+1=-1.$
(6)$\frac{2}{5}-\left|-1 \frac{1}{2}\right|-(+2 \frac{1}{4})-(-2.75)$
$=\frac{2}{5}-1 \frac{1}{2}-2 \frac{1}{4}+2.75=\frac{2}{5}+(-1 \frac{1}{2}-2 \frac{1}{4}+2.75)$
$=\frac{2}{5}-1=-\frac{3}{5}.$
$=(-20)+(+12)+(+5)+(-7)=-10.$
(2)$3 \frac{1}{2}+(-\frac{1}{2})-(-\frac{1}{3})+2 \frac{2}{3}$
$=(3 \frac{1}{2}-\frac{1}{2})+(\frac{1}{3}+2 \frac{2}{3})=3+3=6.$
(3)$5.6-0.9+4.4+(-8.1)-(+0.1)$
$=(5.6+4.4)+(-0.9-8.1-0.1)$
$=10-9.1=0.9.$
(4)$(+3.5)-(+\frac{2}{5})-(-6.5)+(-\frac{3}{5})$
$=3.5-\frac{2}{5}+6.5-\frac{3}{5}$
$=3.5+6.5-\frac{2}{5}-\frac{3}{5}$
$=10-1=9.$
(5)$(-3 \frac{2}{3})-(-2 \frac{3}{4})-(-1 \frac{2}{3})-(+1.75)$
$=-3 \frac{2}{3}+2 \frac{3}{4}+1 \frac{2}{3}+(-1 \frac{3}{4})$
$=(-3 \frac{2}{3}+1 \frac{2}{3})+[(+2 \frac{3}{4})+(-1 \frac{3}{4})]$
$=-2+1=-1.$
(6)$\frac{2}{5}-\left|-1 \frac{1}{2}\right|-(+2 \frac{1}{4})-(-2.75)$
$=\frac{2}{5}-1 \frac{1}{2}-2 \frac{1}{4}+2.75=\frac{2}{5}+(-1 \frac{1}{2}-2 \frac{1}{4}+2.75)$
$=\frac{2}{5}-1=-\frac{3}{5}.$
7.「2026河北石家庄月考，★☆」从-2中减去$-\frac{5}{12}$与$-\frac{1}{8}$的和，所得的差是（
A.$-1\frac{11}{24}$
B.$-\frac{7}{24}$
C.$-2\frac{7}{24}$
D.$\frac{7}{24}$
A
）A.$-1\frac{11}{24}$
B.$-\frac{7}{24}$
C.$-2\frac{7}{24}$
D.$\frac{7}{24}$
答案
7.A $-2-[(-\frac{5}{12})+(-\frac{1}{8})]=-2-(-\frac{13}{24})=-1 \frac{11}{24}.$故选 A.
