2026年配套练习与检测五年级数学下册北师大版第15页答案
1. 计算下列各题。(能简算的要简算)
$ \frac{1}{3} + \frac{1}{6} - \frac{1}{2} $
$ \frac{4}{5} + \frac{3}{7} + \frac{1}{5} $
$ \frac{1}{8} + \frac{5}{9} + \frac{3}{8} + \frac{4}{9} $
$ \frac{5}{7} - (\frac{4}{7} + \frac{1}{9}) $
$ \frac{2}{5} - \frac{1}{6} + \frac{3}{5} - \frac{5}{6} $
$ \frac{9}{5} - \frac{2}{7} - \frac{5}{7} $

答案

1. $\frac{1}{3} + \frac{1}{6} - \frac{1}{2}$
解:
$\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$
$\frac{1}{2} - \frac{1}{2} = 0$
答案:0
2. $\frac{4}{5} + \frac{3}{7} + \frac{1}{5}$
解:
$\frac{4}{5} + \frac{1}{5} = 1$
$1 + \frac{3}{7} = 1\frac{3}{7}$
答案:$1\frac{3}{7}$
3. $\frac{1}{8} + \frac{5}{9} + \frac{3}{8} + \frac{4}{9}$
解:
$(\frac{1}{8} + \frac{3}{8}) + (\frac{5}{9} + \frac{4}{9}) = \frac{4}{8} + 1 = \frac{1}{2} + 1 = 1\frac{1}{2}$
答案:$1\frac{1}{2}$
4. $\frac{5}{7} - (\frac{4}{7} + \frac{1}{9})$
解:
$\frac{5}{7} - \frac{4}{7} - \frac{1}{9} = \frac{1}{7} - \frac{1}{9} = \frac{9}{63} - \frac{7}{63} = \frac{2}{63}$
答案:$\frac{2}{63}$
5. $\frac{2}{5} - \frac{1}{6} + \frac{3}{5} - \frac{5}{6}$
解:
$(\frac{2}{5} + \frac{3}{5}) - (\frac{1}{6} + \frac{5}{6}) = 1 - 1 = 0$
答案:0
6. $\frac{9}{5} - \frac{2}{7} - \frac{5}{7}$
解:
$\frac{9}{5} - (\frac{2}{7} + \frac{5}{7}) = \frac{9}{5} - 1 = \frac{4}{5}$
答案:$\frac{4}{5}$
2. 求下列图形的表面积。
(1)

(2)

答案

(1)$236\,\mathrm{dm}^2$;(2)$216\,\mathrm{m}^2$

解析

(1)长方体表面积公式:$S=(ab+ah+bh)×2$,其中$a=8\,\mathrm{dm}$,$b=6\,\mathrm{dm}$,$h=5\,\mathrm{dm}$。
$S=(8×6 + 8×5 + 6×5)×2$
$=(48 + 40 + 30)×2$
$=118×2$
$=236\,\mathrm{dm}^2$
(2)正方体表面积公式:$S=6a^2$,其中$a=6\,\mathrm{m}$。
$S=6×6^2$
$=6×36$
$=216\,\mathrm{m}^2$