《2024年基础训练大象出版社七年级数学上册人教版》

1. ($★★$)计算：
(1)$-1^{2}-\left(-\dfrac {1}{2}\right)^{3}÷ 4$；
(2)$-4^{3}+8× (-1)^{2024}-12÷ \left(-\dfrac {2}{3}\right)$；
(3)$-3^{2}× \left(-\dfrac {1}{3}\right)^{2}-(-2)^{3}÷ \left(-\dfrac {1}{2}\right)^{2}$；
(4)$(-2)^{4}÷ (-8)-\left(-\dfrac {1}{2}\right)^{3}× (-2^{2})$；
(5)$\left(-\dfrac {5}{8}\right)× (-4)^{2}-0.25× (-5)× (-4)^{3}$；
(6)$(-4^{2})÷ \left(2\dfrac {2}{3}\right)^{2}+5\dfrac {1}{2}× \left(-\dfrac {1}{6}\right)-(-0.5)^{2}$；
(7)$\dfrac {2^{2}}{3}+(-3^{2}+5)+(-3)^{2}× \left(\dfrac {2}{3}\right)^{2}$；
(8)$(-2)^{3}+(-3)× [(-4)^{2}+2]-(-3)^{2}÷ (-2)$；
(9)$-1^{4}+(1-0.5)× \dfrac {1}{3}× [2-(-3)^{2}]$；
(10)$-(-1)^{4}-\left[1\dfrac {3}{7}+(-12)÷ 6\right]^{2}× (-7)^{3}+|-2-1|$。

答案： 解:
(1)原式$=-1-(-\frac{1}{8})×\frac{1}{4}$
$=-1+\frac{1}{32}$
$=-\frac{31}{32}$
解:
(2)原式$=-64+8-12×(-\frac{3}{2})$
=-64+8+18
=-56+18
=-38
解:
(3)原式$=-9×\frac{1}{9}-(-8)×4$
=-1+32
=31

解:
(4)原式$=16÷(-8)-(-\frac{1}{8})×(-4)$
$=-2-\frac{1}{2}$
$=-2\frac{1}{2}$
解:
(5)原式$=-\frac {5}{8}×16-0.25×(-5)×(-64)$
=-10-80
=-90
解:
(6)原式$=-16÷(\frac{8}{3})²+\frac{11}{2}×(-\frac{1}{6})-\frac{1}{4}$
$=-16×\frac{9}{64}-\frac {11}{12}-\frac {1}{4}$
$=-\frac{9}{4}-\frac{1}{4}-\frac {11}{12}$
$=-\frac{41}{12}$
解:
(7)原式$=\frac{4}{3}+(-9+5)+9×\frac{4}{9}$
$=\frac{4}{3}-4+4$
$=\frac{4}{3}$
解:
(8)原式=-8+(-3)×(16+2)-9÷(-2)
$=-8-3×18+\frac{9}{2}$
=-8-54+4.5
=-62+4.5
=-57.5
解:
(9)原式$=-1+\frac{1}{2}×\frac{1}{3}×(2-9)$
$=-1+\frac{1}{6}×(-7)$
$=-1-\frac{7}{6}$
$=-\frac{13}{6}$
解:
(10)原式$=-1-(\frac {10}{7}-2)²×(-7)^3+3$
$=-1-\frac{16}{49}×(7)^3+3$
=-1+16×7+3
=-1+112+3
=114.