《2025年绩优学案七年级数学上册人教版》

5. 计算：
(1) $ 1.125 + \left(-3\dfrac{2}{5}\right) - \dfrac{1}{8} + (-0.6) $；
(2) $ \left(-\dfrac{5}{6}\right) ÷ (-3) × \left(-1\dfrac{4}{5}\right) × (-2) $；
(3) $ -3^2 - (-5)^3 × \left(\dfrac{2}{5}\right)^2 - 15 ÷ |-3| $；
(4) $ -5^2 × \left(-\dfrac{1}{5}\right)^2 + \left(\dfrac{3}{4} - \dfrac{1}{6} + \dfrac{3}{8}\right) × (-24) $；
(5) $ -2^2 ÷ \dfrac{4}{3} - \left[2^2 - \left(1 - \dfrac{1}{2} × \dfrac{1}{3}\right)\right] × 12 $。

答案：
(1) $1.125 + \left(-3\dfrac{2}{5}\right) - \dfrac{1}{8} + (-0.6)$
$=1.125 - \dfrac{1}{8} + \left(-3\dfrac{2}{5} - 0.6\right)$
$=1.125 - 0.125 + (-3.4 - 0.6)$
$=1 + (-4)$
$=-3$
(2) $\left(-\dfrac{5}{6}\right) ÷ (-3) × \left(-1\dfrac{4}{5}\right) × (-2)$
$=\left(-\dfrac{5}{6}\right) × \left(-\dfrac{1}{3}\right) × \left(-\dfrac{9}{5}\right) × (-2)$
$=\left(\dfrac{5}{6} × \dfrac{9}{5}\right) × \left(\dfrac{1}{3} × 2\right)$
$=\dfrac{3}{2} × \dfrac{2}{3}$
$=1$
(3) $-3^2 - (-5)^3 × \left(\dfrac{2}{5}\right)^2 - 15 ÷ |-3|$
$=-9 - (-125) × \dfrac{4}{25} - 15 ÷ 3$
$=-9 - (-20) - 5$
$=-9 + 20 - 5$
$=6$
(4) $-5^2 × \left(-\dfrac{1}{5}\right)^2 + \left(\dfrac{3}{4} - \dfrac{1}{6} + \dfrac{3}{8}\right) × (-24)$
$=-25 × \dfrac{1}{25} + \left[\dfrac{3}{4}×(-24) - \dfrac{1}{6}×(-24) + \dfrac{3}{8}×(-24)\right]$
$=-1 + (-18 + 4 - 9)$
$=-1 + (-23)$
$=-24$
(5) $-2^2 ÷ \dfrac{4}{3} - \left[2^2 - \left(1 - \dfrac{1}{2} × \dfrac{1}{3}\right)\right] × 12$
$=-4 × \dfrac{3}{4} - \left[4 - \left(1 - \dfrac{1}{6}\right)\right] × 12$
$=-3 - \left[4 - \dfrac{5}{6}\right] × 12$
$=-3 - \dfrac{19}{6} × 12$
$=-3 - 38$
$=-41$