2026年课时提优计划作业本七年级数学上册苏科版第52页答案
1. 计算:
(1)$12+(-2)+(-5)+8$;
(2)$-\dfrac{1}{2}-(+\dfrac{5}{4})-(-\dfrac{3}{2})+(-\dfrac{1}{4})$;
(3)$(-81)÷2\dfrac{1}{4}×(-\dfrac{4}{9})÷6$;
(4)$-15×\dfrac{2}{3}+(-\dfrac{3}{4})÷0.125$;
(5)$1.25×(-3.2)÷(0.5-\dfrac{2}{3})÷2\dfrac{2}{3}$;
(6)$1+[20-(-8)]÷(-4)$;
(7)$(1-1\dfrac{1}{2}-\dfrac{3}{8}+\dfrac{7}{12})×(-24)$;
(8)$72×\dfrac{2}{9}+72÷1.5$;
(9)$9\dfrac{4}{5}×(-5)$;
(10)$3-(\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{3}{4})×(-24)$。
2. 计算:
(1)$(-1)^4-\dfrac{1}{6}×[2-(-3)^2]$;
(2)$(-1\dfrac{1}{4})×\dfrac{2}{5}×8-9÷(-\dfrac{3}{2})^2$;
(3)$-1^{2\ 026}-(2-0.5)×\dfrac{1}{3}×[1-(-3)^2]$;
(4)$-1^4-[(1-0.7)×\dfrac{1}{3}]×[3-(-2)^2]$;
(5)$-3^2×(-\dfrac{1}{3})^2+(-\dfrac{3}{4}-\dfrac{1}{6}+\dfrac{3}{8})×(-24)$;
(6)$-3-[-2^2+(2^3-4)÷(1-1\dfrac{1}{3})]$;
(7)$5-(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{2}{5})÷(-\dfrac{1}{30})$;
(8)$-9^2+2×(-3)^2+(-6)÷(-\dfrac{2}{3})^2$;
(9)$(-3)^2×[-\dfrac{2}{3}+(-\dfrac{5}{6})]-6÷(-2)×(-\dfrac{1}{3})$;
(10)$-2^3÷\dfrac{4}{3}×(\dfrac{1}{2}-\dfrac{1}{3})^2+[3-(-1)^3]$。

答案

1. (1)原式$=(12+8)+[(-2)+(-5)]=20-7=13$.
(2)原式$=-\dfrac{1}{2}+(-\dfrac{5}{4})+\dfrac{3}{2}+(-\dfrac{1}{4})=(-\dfrac{1}{2}+\dfrac{3}{2})+[(-\dfrac{5}{4})+(-\dfrac{1}{4})]=1+(-\dfrac{3}{2})=-\dfrac{1}{2}$.
(3)原式$=81÷\dfrac{9}{4}×\dfrac{4}{9}÷6=81×\dfrac{4}{9}×\dfrac{4}{9}×\dfrac{1}{6}=\dfrac{8}{3}$.
(4)原式$=-10+(-\dfrac{3}{4})×8=-10-6=-16$.
(5)原式$=\dfrac{5}{4}×(-\dfrac{16}{5})÷(-\dfrac{1}{6})÷\dfrac{8}{3}=\dfrac{5}{4}×\dfrac{16}{5}×6×\dfrac{3}{8}=9$.
(6)原式$=1+28÷(-4)=1+(-7)=-6$.
(7)原式$=-24-\dfrac{3}{2}×(-24)-\dfrac{3}{8}×(-24)+\dfrac{7}{12}×(-24)=-24+36+9-14=7$.
(8)原式$=16+72÷\dfrac{3}{2}=16+72×\dfrac{2}{3}=16+48=64$.
(9)原式$=(10-\dfrac{1}{5})×(-5)=10×(-5)-\dfrac{1}{5}×(-5)=-50+1=-49$.
(10)原式$=3-[\dfrac{1}{3}×(-24)-\dfrac{1}{6}×(-24)-\dfrac{3}{4}×(-24)]=3-(-8+4+18)=3-14=-11$.
2. (1)原式$=1-\dfrac{1}{6}×(2-9)=1-\dfrac{1}{6}×(-7)=1+\dfrac{7}{6}=\dfrac{13}{6}$.
(2)原式$=-\dfrac{5}{4}×\dfrac{2}{5}×8-9÷\dfrac{9}{4}=-4-9×\dfrac{4}{9}=-4-4=-8$.
(3)原式$=-1-\dfrac{3}{2}×\dfrac{1}{3}×(-8)=-1+4=3$.
(4)原式$=-1-\dfrac{1}{10}×(-1)=-1+\dfrac{1}{10}=-\dfrac{9}{10}$.
(5)原式$=-9×\dfrac{1}{9}+(-\dfrac{3}{4})×(-24)-\dfrac{1}{6}×(-24)+\dfrac{3}{8}×(-24)=-1+18+4-9=12$.
(6)原式$=-3-[-4+4÷(-\dfrac{1}{3})]=-3-(-4-12)=-3+16=13$.
(7)原式$=5-(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{2}{5})×(-30)=5-[\dfrac{1}{2}×(-30)-\dfrac{2}{3}×(-30)+\dfrac{2}{5}×(-30)]=5-(-15+20-12)=5+15-20+12=12$.
(8)原式$=-81+2×9+(-6)÷\dfrac{4}{9}=-81+18-6×\dfrac{9}{4}=-63-\dfrac{27}{2}=-\dfrac{153}{2}$.
(9)原式$=9×(-\dfrac{3}{2})-(-3)×(-\dfrac{1}{3})=-\dfrac{27}{2}-1=-\dfrac{29}{2}$.
(10)原式$=-8×\dfrac{3}{4}×\dfrac{1}{36}+4=-\dfrac{1}{6}+4=3\dfrac{5}{6}$.