2026年拔尖特训七年级数学上册人教版第31页答案
1. 计算:
(1) $-101+36+(-12)+(-37)-(-28)$.
(2) $-1\dfrac{2}{3}-2\dfrac{3}{5}+5-\dfrac{1}{3}-\dfrac{2}{5}+4$.

答案

(1) 原式$=(-101-12-37)+(36+28)=-150+64=-86$.
(2) 原式$=(-1\dfrac{2}{3}-\dfrac{1}{3})+(-2\dfrac{3}{5}-\dfrac{2}{5})+(5+4)=-2-3+9=4$.
2. 计算:
(1) $-20.75 - 3.25 + \frac{1}{4} + 19\frac{3}{4}$.
(2) $(-5\frac{1}{3}) + 4.19 + 5\frac{3}{8} + (-4\frac{3}{4}) + 9.81 + (-4\frac{2}{3})$.

答案

(1) 原式$=(-20.75+19\dfrac{3}{4})+(\dfrac{1}{4}-3.25)=-1-3=-4$.
(2) 原式$=(-5\dfrac{1}{3}-4\dfrac{2}{3})+(4.19+9.81)+(5\dfrac{3}{8}-4\dfrac{3}{4})=-10+14+\dfrac{5}{8}=4\dfrac{5}{8}$.
3. 计算:
(1) $-5\frac{5}{6} + (-9\frac{2}{3}) + 17\frac{3}{4} + (-3\frac{1}{2})$。
(2) $(-4\frac{7}{8}) + (+8\frac{1}{4}) + (-3\frac{1}{8})$。
(3) $(-99\frac{48}{49}) × (-7)$。
(4) $-132\frac{11}{16} ÷ (-11)$。

答案

(1) 原式$=[(-5)+(-\dfrac{5}{6})]+[(-9)+(-\dfrac{2}{3})]+(17+\dfrac{3}{4})+[(-3)+(-\dfrac{1}{2})]=[(-5)+(-9)+17+(-3)]+[(-\dfrac{5}{6})+(-\dfrac{2}{3})+\dfrac{3}{4}+(-\dfrac{1}{2})]=0+(-\dfrac{5}{4})=-\dfrac{5}{4}$.
(2) 原式$=[(-4)+(-\dfrac{7}{8})]+(8+\dfrac{1}{4})+[(-3)+(-\dfrac{1}{8})]=(-4+8-3)+(-\dfrac{7}{8}+\dfrac{1}{4}-\dfrac{1}{8})=1+(-\dfrac{3}{4})=\dfrac{1}{4}$.
(3) 原式$=(100-\dfrac{1}{49})× 7=700-\dfrac{1}{7}=699\dfrac{6}{7}$.
(4) 原式$=(132+\dfrac{11}{16})× \dfrac{1}{11}=12+\dfrac{1}{16}=12\dfrac{1}{16}$.