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初中数学
六年级
上册
鲁教版
1.「2026 山东青岛黄岛月考节选」计算：$(-3 \dfrac{2}{3})-(-2.75)+3 \dfrac{2}{3}+(+5.75)$
六年级
上册
鲁教版
1.「2026 山东青岛黄岛月考节选」计算：$(-3 \dfrac{2}{3})-(-2.75)+3 \dfrac{2}{3}+(+5.75)$
答案
1.解析 原式$=(-3 \dfrac{2}{3}+3 \dfrac{2}{3})+[5.75-(-2.75)]$
$=0+8.5=8.5.$
$=0+8.5=8.5.$
2.计算：$(-1.25)×\dfrac{5}{7}×(-4)×(-\dfrac{7}{5}).$
答案
2.解析 原式$=[(-1.25)×(-4)]×[\dfrac{5}{7}×(-\dfrac{7}{5})]=5×(-1)=-5.$
3.「2026 山东泰安岱岳期末」计算：
$-8.9 -5\frac{2}{3} +6.7 +2\frac{2}{3} +8.9 -1.7$
$-8.9 -5\frac{2}{3} +6.7 +2\frac{2}{3} +8.9 -1.7$
答案
3.解析 $-8.9-5\dfrac{2}{3}+6.7+2\dfrac{2}{3}+8.9-1.7$
$=(-8.9+8.9)+(-5\dfrac{2}{3}+2\dfrac{2}{3})+(6.7-1.7)$
$=0+(-3)+5=2.$
$=(-8.9+8.9)+(-5\dfrac{2}{3}+2\dfrac{2}{3})+(6.7-1.7)$
$=0+(-3)+5=2.$
4.计算：
$3\frac{7}{12}+(-1\frac{1}{4})+(-3\frac{7}{12})+1\frac{1}{4}+(-4\frac{1}{8})-(-4\frac{1}{8}).$
$3\frac{7}{12}+(-1\frac{1}{4})+(-3\frac{7}{12})+1\frac{1}{4}+(-4\frac{1}{8})-(-4\frac{1}{8}).$
答案
4.解析 $3\dfrac{7}{12}+(-1\dfrac{1}{4})+(-3\dfrac{7}{12})+1\dfrac{1}{4}+(-4\dfrac{1}{8})-(-4\dfrac{1}{8})$
$=(3\dfrac{7}{12}-3\dfrac{7}{12})+(-1\dfrac{1}{4}+1\dfrac{1}{4})+(-4\dfrac{1}{8}+4\dfrac{1}{8})$
$=0+0+0=0.$
$=(3\dfrac{7}{12}-3\dfrac{7}{12})+(-1\dfrac{1}{4}+1\dfrac{1}{4})+(-4\dfrac{1}{8}+4\dfrac{1}{8})$
$=0+0+0=0.$
5.计算:$0.7×1\frac{4}{9}-15×2\frac{3}{4}+\frac{5}{9}×0.7-\frac{1}{4}×15.$
答案
5.解析 $0.7×1\dfrac{4}{9}-15×2\dfrac{3}{4}+\dfrac{5}{9}×0.7-\dfrac{1}{4}×15$
$=0.7×(1\dfrac{4}{9}+\dfrac{5}{9})-15×(2\dfrac{3}{4}+\dfrac{1}{4})$
$=0.7×2-15×3$
$=1.4-45$
$=-43.6.$
$=0.7×(1\dfrac{4}{9}+\dfrac{5}{9})-15×(2\dfrac{3}{4}+\dfrac{1}{4})$
$=0.7×2-15×3$
$=1.4-45$
$=-43.6.$
6.计算：$(-\dfrac{1}{24})÷(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{3}{8}).$
答案
6.解析 原式的倒数为$(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{3}{8})÷(-\dfrac{1}{24})=(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{3}{8})×(-24)=-8+4-9=-13,$
则$(-\dfrac{1}{24})÷(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{3}{8})=-\dfrac{1}{13}.$
则$(-\dfrac{1}{24})÷(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{3}{8})=-\dfrac{1}{13}.$
7.计算：$3\frac{5}{9}+(-9\frac{2}{3})+7\frac{1}{2}+(-1\frac{3}{4}).$
答案
7.解析 $3\dfrac{5}{9}+(-9\dfrac{2}{3})+7\dfrac{1}{2}+(-1\dfrac{3}{4})$
$=(3+\dfrac{5}{9})+[(-9)+(-\dfrac{2}{3})]+(7+\dfrac{1}{2})+[(-1)+(-\dfrac{3}{4})]$
$=[3+(-9)+7+(-1)]+[\dfrac{5}{9}+(-\dfrac{2}{3})]+[\dfrac{1}{2}+(-\dfrac{3}{4})]$
$=0-\dfrac{1}{9}+(-\dfrac{1}{4})=-\dfrac{13}{36}.$
$=(3+\dfrac{5}{9})+[(-9)+(-\dfrac{2}{3})]+(7+\dfrac{1}{2})+[(-1)+(-\dfrac{3}{4})]$
$=[3+(-9)+7+(-1)]+[\dfrac{5}{9}+(-\dfrac{2}{3})]+[\dfrac{1}{2}+(-\dfrac{3}{4})]$
$=0-\dfrac{1}{9}+(-\dfrac{1}{4})=-\dfrac{13}{36}.$
8.计算：$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{2024×2025}$
答案
8.解析 $\dfrac{1}{1×2}+\dfrac{1}{2×3}+\dfrac{1}{3×4}+\dots+\dfrac{1}{2024×2025}$
$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{2024}-\dfrac{1}{2025}$
$=1-\dfrac{1}{2025}=\dfrac{2024}{2025}.$
$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{2024}-\dfrac{1}{2025}$
$=1-\dfrac{1}{2025}=\dfrac{2024}{2025}.$
