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1.$\frac{2}{3}+(-2.5)+3.5+(-\frac{2}{3})=[\frac{2}{3}+(-\frac{2}{3})]+[(-2.5)+3.5]$这个运算中运用了 （
A.加法的交换律
B.加法的结合律
C.加法的交换律和结合律
D.无法确定
C
）A.加法的交换律
B.加法的结合律
C.加法的交换律和结合律
D.无法确定
答案
1.C
2.(东海县月考)下列变形,运用运算律正确的是（
A.$2+(-1)=1+2$
B.$3+(-2)+5=(-2)+3+5$
C.$[6+(-3)]+5=[6+(-5)]+3$
D.$\dfrac{1}{3}+(-2)+(+\dfrac{2}{3})=(\dfrac{1}{3}+\dfrac{2}{3})+(+2)$
B
）A.$2+(-1)=1+2$
B.$3+(-2)+5=(-2)+3+5$
C.$[6+(-3)]+5=[6+(-5)]+3$
D.$\dfrac{1}{3}+(-2)+(+\dfrac{2}{3})=(\dfrac{1}{3}+\dfrac{2}{3})+(+2)$
答案
2.B
3.(涟水县月考)计算：$(-\dfrac{2}{7})+(-\dfrac{5}{7})-(-3)=$
2
.答案
3.2
4.(靖江月考)比$-3\dfrac{1}{2}$大而比$2\dfrac{1}{3}$小的所有整数的和为
-3
.答案
4.-3
5.计算：
(1)$5.6+4.4+(-8.1)$；
(2)$(-7)+(-4)+(+9)+(-5)$；
(3)$\dfrac{1}{4}+(-\dfrac{2}{3})+\dfrac{5}{6}+(-\dfrac{1}{4})+(-\dfrac{1}{3})$；
(4)$5\dfrac{3}{5}+(-5\dfrac{2}{3})+4\dfrac{2}{5}+(-\dfrac{1}{3}).$
(1)$5.6+4.4+(-8.1)$；
(2)$(-7)+(-4)+(+9)+(-5)$；
(3)$\dfrac{1}{4}+(-\dfrac{2}{3})+\dfrac{5}{6}+(-\dfrac{1}{4})+(-\dfrac{1}{3})$；
(4)$5\dfrac{3}{5}+(-5\dfrac{2}{3})+4\dfrac{2}{5}+(-\dfrac{1}{3}).$
答案
5.解:(1)原式=10+(-8.1)=1.9.
(2)原式=(-7)+(-4)+(-5)+(+9)=-16+9=-7.
(3)原式=$(\frac{1}{4}-\frac{1}{4})+[(-\frac{2}{3})+(-\frac{1}{3})]+\frac{5}{6}=$
$0+(-1)+\frac{5}{6}=-\frac{1}{6}.$
(4)原式=$(5\frac{3}{5}+4\frac{2}{5})+[(-5\frac{2}{3})+(-\frac{1}{3})]=$
$10+(-6)=4.$
(2)原式=(-7)+(-4)+(-5)+(+9)=-16+9=-7.
(3)原式=$(\frac{1}{4}-\frac{1}{4})+[(-\frac{2}{3})+(-\frac{1}{3})]+\frac{5}{6}=$
$0+(-1)+\frac{5}{6}=-\frac{1}{6}.$
(4)原式=$(5\frac{3}{5}+4\frac{2}{5})+[(-5\frac{2}{3})+(-\frac{1}{3})]=$
$10+(-6)=4.$
6. 下列各式中,运用加法结合律变形错误的是 （
A.$1+(-0.25)+(-0.75)=1+[(-0.25)+(-0.75)]$
B.$1-2+3-4+5-6=(1-2)+(3-4)+(5-6)$
C.$\dfrac{3}{4}-\dfrac{1}{6}-\dfrac{1}{2}+\dfrac{2}{3}=(\dfrac{3}{4}+\dfrac{1}{2})+(-\dfrac{1}{6}+\dfrac{2}{3})$
D.$7-8-3+6+2=(7-3)+(-8)+(6+2)$
C
）A.$1+(-0.25)+(-0.75)=1+[(-0.25)+(-0.75)]$
B.$1-2+3-4+5-6=(1-2)+(3-4)+(5-6)$
C.$\dfrac{3}{4}-\dfrac{1}{6}-\dfrac{1}{2}+\dfrac{2}{3}=(\dfrac{3}{4}+\dfrac{1}{2})+(-\dfrac{1}{6}+\dfrac{2}{3})$
D.$7-8-3+6+2=(7-3)+(-8)+(6+2)$
答案
6.C
