2025年同步练习西南大学出版社六年级数学上册西师大版四川专版


注:目前有些书本章节名称可能整理的还不是很完善,但都是按照顺序排列的,请同学们按照顺序仔细查找。练习册 2025年同步练习西南大学出版社六年级数学上册西师大版四川专版 答案主要是用来给同学们做完题方便对答案用的,请勿直接抄袭。



《2025年同步练习西南大学出版社六年级数学上册西师大版四川专版》

1. 口算。
$\dfrac{3}{5}÷3=$
$\dfrac{1}{5}$
$\dfrac{3}{7}×2=$
$\dfrac{6}{7}$
$\dfrac{1}{2}÷3=$
$\dfrac{1}{6}$
$3÷\dfrac{3}{5}=$
5

$\dfrac{2}{5}÷\dfrac{1}{5}=$
2
$\dfrac{1}{4}÷\dfrac{2}{3}=$
$\dfrac{3}{8}$
$\dfrac{1}{3}÷\dfrac{1}{2}=$
$\dfrac{2}{3}$
$6×\dfrac{1}{3}=$
2
答案: $\dfrac{3}{5}÷3=\dfrac{1}{5}$;$\dfrac{3}{7}×2=\dfrac{6}{7}$;$\dfrac{1}{2}÷3=\dfrac{1}{6}$;$3÷\dfrac{3}{5}=5$;$\dfrac{2}{5}÷\dfrac{1}{5}=2$;$\dfrac{1}{4}÷\dfrac{2}{3}=\dfrac{3}{8}$;$\dfrac{1}{3}÷\dfrac{1}{2}=\dfrac{2}{3}$;$6×\dfrac{1}{3}=2$
2. 在$◯$里填“$>$”或“$<$”。
$\dfrac{6}{7}÷3◯\dfrac{6}{7}$ $\dfrac{6}{7}÷\dfrac{3}{4}◯\dfrac{6}{7}$
$\dfrac{6}{7}×3◯\dfrac{6}{7}$ $\dfrac{6}{7}×\dfrac{3}{4}◯\dfrac{6}{7}$
<;>;>;<
答案: $\dfrac{6}{7}÷3=\dfrac{6}{7}×\dfrac{1}{3}=\dfrac{2}{7}$,因为$\dfrac{2}{7}<\dfrac{6}{7}$,所以$\dfrac{6}{7}÷3<\dfrac{6}{7}$;
$\dfrac{6}{7}÷\dfrac{3}{4}=\dfrac{6}{7}×\dfrac{4}{3}=\dfrac{8}{7}$,因为$\dfrac{8}{7}>\dfrac{6}{7}$,所以$\dfrac{6}{7}÷\dfrac{3}{4}>\dfrac{6}{7}$;
$\dfrac{6}{7}×3=\dfrac{18}{7}$,因为$\dfrac{18}{7}>\dfrac{6}{7}$,所以$\dfrac{6}{7}×3>\dfrac{6}{7}$;
$\dfrac{6}{7}×\dfrac{3}{4}=\dfrac{9}{14}$,$\dfrac{6}{7}=\dfrac{12}{14}$,因为$\dfrac{9}{14}<\dfrac{12}{14}$,所以$\dfrac{6}{7}×\dfrac{3}{4}<\dfrac{6}{7}$。
$<$;$>$;$>$;$<$
3. 计算。
$\dfrac{2}{3}×\dfrac{3}{5}÷\dfrac{7}{10}$ $\dfrac{8}{9}÷\dfrac{5}{6}×\dfrac{5}{8}$
$21÷\dfrac{7}{8}÷\dfrac{3}{4}$ $7×\dfrac{5}{28}÷\dfrac{9}{10}$
答案: 1.
$\dfrac{2}{3} × \dfrac{3}{5} ÷ \dfrac{7}{10}$
$=\dfrac{2}{3} × \dfrac{3}{5} × \dfrac{10}{7}$
$=\dfrac{2}{5} × \dfrac{10}{7}$
$=\dfrac{4}{7}$
2.
$\dfrac{8}{9} ÷ \dfrac{5}{6} × \dfrac{5}{8}$
$=\dfrac{8}{9} × \dfrac{6}{5} × \dfrac{5}{8}$
$=\dfrac{8 × 6 × 5}{9 × 5 × 8}$
$=\dfrac{2}{3}$
3.
$21 ÷ \dfrac{7}{8} ÷ \dfrac{3}{4}$
$=21 × \dfrac{8}{7} × \dfrac{4}{3}$
$=24 × \dfrac{4}{3}$
$=32$
4.
$7 × \dfrac{5}{28} ÷ \dfrac{9}{10}$
$=\dfrac{35}{28} × \dfrac{10}{9}$
$=\dfrac{5}{4} × \dfrac{10}{9}$
$=\dfrac{25}{18}$

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