《2025年计算高手六年级数学上册苏教版》

一、直接写出得数。
$\frac {6}{7}÷\frac {2}{21}= $
$\frac {8}{11}÷\frac {4}{5}= $
$\frac {3}{5}÷\frac {9}{10}= $
$\frac {8}{27}÷\frac {4}{9}= $
$\frac {3}{11}÷\frac {9}{22}= $
$\frac {8}{27}÷\frac {4}{5}= $
$\frac {3}{23}÷\frac {3}{46}= $
$\frac {6}{7}÷\frac {3}{14}= $
$27÷\frac {9}{7}= $
$15÷\frac {5}{8}= $
$\frac {2}{15}÷\frac {3}{5}= $
$16÷\frac {4}{13}= $
$\frac {15}{16}÷\frac {5}{4}= $
$\frac {1}{6}÷\frac {4}{9}= $
$18÷\frac {12}{11}= $

答案： 9
$\frac {10}{11}$
$\frac {2}{3}$
$\frac {2}{3}$
$\frac {2}{3}$
$\frac {10}{27}$
2
4
21
24
$\frac {2}{9}$
52
$\frac {3}{4}$
$\frac {3}{8}$
$\frac {33}{2}$

二、脱式计算。
$\frac {7}{9}÷\frac {2}{3}÷\frac {7}{9}$
$\frac {3}{5}×\frac {2}{9}×15$
$\frac {1}{7}÷\frac {2}{3}×7$

答案： 计算$\frac {7}{9}÷\frac {2}{3}÷\frac {7}{9}$
$\frac {7}{9}÷\frac {2}{3}÷\frac {7}{9}=\frac {7}{9}÷\frac {7}{9}÷\frac {2}{3}$
$ = 1÷\frac{2}{3}$
$=1×\frac{3}{2}=\frac{3}{2}$
计算$\frac {3}{5}×\frac {2}{9}×15$
$\frac {3}{5}×\frac {2}{9}×15=\frac {3}{5}×15×\frac {2}{9}$
$=9×\frac{2}{9}$
$ = 2$
计算$\frac {1}{7}÷\frac {2}{3}×7$
$\frac {1}{7}÷\frac {2}{3}×7=\frac {1}{7}×\frac {3}{2}×7$
$\frac {1}{7}×\frac {3}{2}×7=\frac {1}{7}×7×\frac {3}{2}$
$=1×\frac{3}{2}$
$=\frac{3}{2}$
综上，答案依次为$\boldsymbol{\frac{3}{2}}$；$\boldsymbol{2}$；$\boldsymbol{\frac{3}{2}}$ 。

三、在〇里填上“＞”“＜”或“=”。
$\frac {15}{16}×\frac {9}{4}$〇$\frac {15}{16}$
$\frac {7}{10}×\frac {4}{5}$〇$\frac {7}{10}÷\frac {4}{5}$
$\frac {5}{6}÷\frac {2}{3}$〇$\frac {5}{6}$
$\frac {11}{12}×\frac {1}{2}÷\frac {8}{9}$〇$\frac {11}{12}×\frac {9}{8}÷2$

答案： ＞
＜
＞
=

四、跨学科|电压 电鳗和电鲶都具有奇异的放电功能,有水申的“高压线”之称。家中常用的电压为220伏(标准电压),是电鳗能输出的最大电压的$\frac {11}{40}$。电鲶能输出的最大电压是电鳗能输出的最大电压的$\frac {9}{16}$。电鲶能输出的最大电压是多少伏？

答案： $220÷\frac {11}{40}×\frac {9}{16}=450$(伏)