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7. 计算 $2\dfrac{1}{7}+(-2\dfrac{3}{4})+4\dfrac{6}{7}+(-7\dfrac{1}{4})$ 时,运算律用得最为恰当的是 （
A.$[2\dfrac{1}{7}+(-7\dfrac{1}{4})]+[(-2\dfrac{3}{4})+4\dfrac{6}{7}]$
B.$(2\dfrac{1}{7}+4\dfrac{6}{7})+[(-2\dfrac{3}{4})+(-7\dfrac{1}{4})]$
C.$[(-2\dfrac{3}{4})+7\dfrac{1}{4}]+(2\dfrac{1}{7}+4\dfrac{6}{7})$
D.$(-7\dfrac{1}{4}+2\dfrac{3}{4})+(2\dfrac{1}{7}+4\dfrac{6}{7})$
B
）A.$[2\dfrac{1}{7}+(-7\dfrac{1}{4})]+[(-2\dfrac{3}{4})+4\dfrac{6}{7}]$
B.$(2\dfrac{1}{7}+4\dfrac{6}{7})+[(-2\dfrac{3}{4})+(-7\dfrac{1}{4})]$
C.$[(-2\dfrac{3}{4})+7\dfrac{1}{4}]+(2\dfrac{1}{7}+4\dfrac{6}{7})$
D.$(-7\dfrac{1}{4}+2\dfrac{3}{4})+(2\dfrac{1}{7}+4\dfrac{6}{7})$
答案
7. B
8. 计算 $0.75+(-\dfrac{11}{4})+0.125+(-\dfrac{5}{7})+(-4\dfrac{1}{8})$ 的结果是（
A.$6\dfrac{5}{7}$
B.$-6\dfrac{5}{7}$
C.$5\dfrac{2}{7}$
D.$-5\dfrac{2}{7}$
B
）A.$6\dfrac{5}{7}$
B.$-6\dfrac{5}{7}$
C.$5\dfrac{2}{7}$
D.$-5\dfrac{2}{7}$
答案
8. B 解析：原式$=\dfrac{3}{4}+(-\dfrac{11}{4})+\dfrac{1}{8}+(-\dfrac{5}{7})+(-\dfrac{33}{8})=[\dfrac{3}{4}+(-\dfrac{11}{4})]+[\dfrac{1}{8}+(-\dfrac{33}{8})]+(-\dfrac{5}{7})=-2-4-\dfrac{5}{7}=-6\dfrac{5}{7}$。
9. 分别写出一个含有三个加数且满足条件的等式.
(1)所有的加数都是负数,和是$-13:$
(2)至少有一个加数是正整数,和是$-13:$
(1)所有的加数都是负数,和是$-13:$
$(-1)+(-2)+(-10)=-13$（答案不唯一）
.(2)至少有一个加数是正整数,和是$-13:$
$3+(-3)+(-13)=-13$（答案不唯一）
.答案
9. (1)$(-1)+(-2)+(-10)=-13$（答案不唯一） (2)$3+(-3)+(-13)=-13$（答案不唯一）
10. 计算:$1+(-2)+3+(-4)+5+(-6)+... +99+(-100)=$
$-50$
.答案
10. $-50$ 解析：原式$=(1-2)+(3-4)+(5-6)+\dots+(99-100)=-1-1-1\dots-1=-50$。
11. 计算：
(1)$3\dfrac{7}{12}+(-1\dfrac{1}{4})+(-3\dfrac{7}{12})+1\dfrac{1}{4}+(-4\dfrac{1}{8})$；
(2)$2\dfrac{2}{5}+(-2\dfrac{7}{8})+(-1\dfrac{5}{12})+4\dfrac{3}{5}+(-1)+(-3\dfrac{7}{12}).$
(1)$3\dfrac{7}{12}+(-1\dfrac{1}{4})+(-3\dfrac{7}{12})+1\dfrac{1}{4}+(-4\dfrac{1}{8})$；
(2)$2\dfrac{2}{5}+(-2\dfrac{7}{8})+(-1\dfrac{5}{12})+4\dfrac{3}{5}+(-1)+(-3\dfrac{7}{12}).$
答案
11. (1)原式$=[3\dfrac{7}{12}+(-3\dfrac{7}{12})]+[(-1\dfrac{1}{4})+1\dfrac{1}{4}]+(-4\dfrac{1}{8})=0+0+(-4\dfrac{1}{8})=-4\dfrac{1}{8}$。
(2)原式$=(2\dfrac{2}{5}+4\dfrac{3}{5})+[(-2\dfrac{7}{8})+(-1)] + [(-1\dfrac{5}{12})+(-3\dfrac{7}{12})] =7 + (-3\dfrac{7}{8})+(-5)=7+(-8\dfrac{7}{8})=-1\dfrac{7}{8}$。
(2)原式$=(2\dfrac{2}{5}+4\dfrac{3}{5})+[(-2\dfrac{7}{8})+(-1)] + [(-1\dfrac{5}{12})+(-3\dfrac{7}{12})] =7 + (-3\dfrac{7}{8})+(-5)=7+(-8\dfrac{7}{8})=-1\dfrac{7}{8}$。
12. 阅读第(1)小题的计算方法，再用这种方法计算第(2)小题.
(1)计算：$-5\dfrac{5}{6}+(-9\dfrac{2}{3})+17\dfrac{3}{4}+(-3\dfrac{1}{2}).$
$\begin{aligned}\mathrm{解:原式}&=[(-5)+(-\dfrac{5}{6})]+[(-9)+(-\dfrac{2}{3})]+(17+\dfrac{3}{4})+[(-3)+(-\dfrac{1}{2})]\\&=[(-5)+(-9)+17+(-3)]+[(-\dfrac{5}{6})+(-\dfrac{2}{3})+\dfrac{3}{4}+(-\dfrac{1}{2})]\\&=0+(-1\dfrac{1}{4})=-1\dfrac{1}{4}.\end{aligned}$
上面这种解题方法叫作拆项法.
(2)计算：$(-2\ 026\dfrac{5}{6})+(-2\ 025\dfrac{2}{3})+4\ 052\dfrac{2}{3}+(-1\dfrac{1}{2}).$
(1)计算：$-5\dfrac{5}{6}+(-9\dfrac{2}{3})+17\dfrac{3}{4}+(-3\dfrac{1}{2}).$
$\begin{aligned}\mathrm{解:原式}&=[(-5)+(-\dfrac{5}{6})]+[(-9)+(-\dfrac{2}{3})]+(17+\dfrac{3}{4})+[(-3)+(-\dfrac{1}{2})]\\&=[(-5)+(-9)+17+(-3)]+[(-\dfrac{5}{6})+(-\dfrac{2}{3})+\dfrac{3}{4}+(-\dfrac{1}{2})]\\&=0+(-1\dfrac{1}{4})=-1\dfrac{1}{4}.\end{aligned}$
上面这种解题方法叫作拆项法.
(2)计算：$(-2\ 026\dfrac{5}{6})+(-2\ 025\dfrac{2}{3})+4\ 052\dfrac{2}{3}+(-1\dfrac{1}{2}).$
答案
12. 原式$=(-2\ 026-\dfrac{5}{6})+(-2\ 025-\dfrac{2}{3})+(4\ 052+\dfrac{2}{3})+(-1-\dfrac{1}{2})=(-2\ 026-2\ 025+4\ 052-1)+(-\dfrac{5}{6}-\dfrac{1}{2})+(-\dfrac{2}{3}+\dfrac{2}{3})=0-1\dfrac{1}{3}+0=-1\dfrac{1}{3}$。
