1. 直接写出得数。(20☆)
$\frac{34}{99}×\frac{9}{17}=$
$\frac{5}{13}÷7=$
$\frac{14}{33}+\frac{3}{11}=$
$\frac{49}{50}-\frac{9}{10}=$
$\frac{25}{66}÷\frac{15}{11}=$
$\frac{17}{20}×\frac{10}{51}=$
$1.5÷\frac{15}{23}=$
$\frac{13}{40}×80=$
$\frac{31}{35}-\frac{2}{7}=$
$\frac{26}{45}×\frac{9}{13}=$
$\frac{5}{44}+\frac{3}{11}=$
$\frac{37}{90}÷\frac{37}{30}=$
$\frac{17}{10}+0.3=$
$\frac{27}{40}÷\frac{3}{80}=$
$\frac{49}{55}-\frac{3}{11}=$
$\frac{2}{5}×\frac{7}{24}=$
$1÷\frac{13}{19}=$
$\frac{4}{15}×60=$
$\frac{3}{44}×\frac{33}{8}=$
$\frac{9}{2}÷\frac{3}{20}=$
$\frac{34}{99}×\frac{9}{17}=$
$\frac{5}{13}÷7=$
$\frac{14}{33}+\frac{3}{11}=$
$\frac{49}{50}-\frac{9}{10}=$
$\frac{25}{66}÷\frac{15}{11}=$
$\frac{17}{20}×\frac{10}{51}=$
$1.5÷\frac{15}{23}=$
$\frac{13}{40}×80=$
$\frac{31}{35}-\frac{2}{7}=$
$\frac{26}{45}×\frac{9}{13}=$
$\frac{5}{44}+\frac{3}{11}=$
$\frac{37}{90}÷\frac{37}{30}=$
$\frac{17}{10}+0.3=$
$\frac{27}{40}÷\frac{3}{80}=$
$\frac{49}{55}-\frac{3}{11}=$
$\frac{2}{5}×\frac{7}{24}=$
$1÷\frac{13}{19}=$
$\frac{4}{15}×60=$
$\frac{3}{44}×\frac{33}{8}=$
$\frac{9}{2}÷\frac{3}{20}=$
答案
1. $\frac{2}{11}$
$\frac{5}{91}$
$\frac{23}{33}$
$0.08$
$\frac{5}{18}$
$\frac{1}{6}$
$2.3$
$26$
$\frac{3}{5}$
$\frac{2}{5}$
$\frac{17}{44}$
$\frac{1}{3}$
$2$
$18$
$\frac{34}{55}$
$\frac{7}{60}$
$\frac{19}{13}$
$16$
$\frac{9}{32}$
$30$
$\frac{5}{91}$
$\frac{23}{33}$
$0.08$
$\frac{5}{18}$
$\frac{1}{6}$
$2.3$
$26$
$\frac{3}{5}$
$\frac{2}{5}$
$\frac{17}{44}$
$\frac{1}{3}$
$2$
$18$
$\frac{34}{55}$
$\frac{7}{60}$
$\frac{19}{13}$
$16$
$\frac{9}{32}$
$30$
2. 下面问题能用算式“$40×(1-\dfrac{1}{5})$”解决的是(
A.妈妈购买了一个原价40元的茶壶,元旦促销降价了$\dfrac{1}{5}$,降价了多少元
B.爸爸买了一根绳子,用去了$\dfrac{1}{5}$,还剩下40分米,这根绳子原来有多长
C.小明完成作业需40分钟,现在已经完成了$\dfrac{1}{5}$,要完成剩下的作业还需要多少分钟
C
)。(6☆)A.妈妈购买了一个原价40元的茶壶,元旦促销降价了$\dfrac{1}{5}$,降价了多少元
B.爸爸买了一根绳子,用去了$\dfrac{1}{5}$,还剩下40分米,这根绳子原来有多长
C.小明完成作业需40分钟,现在已经完成了$\dfrac{1}{5}$,要完成剩下的作业还需要多少分钟
答案
2. C
3. 计算下面各题,能简算的要简算。(24☆)
$3÷\frac{3}{8}+\frac{3}{8}÷3$
$27×\frac{3}{10}-\frac{3}{10}÷\frac{1}{7}$
$2-\frac{5}{12}÷\frac{2}{3}-\frac{3}{8}$
$\frac{8}{11}÷[\frac{2}{5}÷(1-\frac{7}{10})]$
$\frac{2}{3}÷\frac{5}{6}×\frac{2}{5}$
$\frac{5}{12}÷6+\frac{1}{6}×\frac{1}{3}$
$3÷\frac{3}{8}+\frac{3}{8}÷3$
$27×\frac{3}{10}-\frac{3}{10}÷\frac{1}{7}$
$2-\frac{5}{12}÷\frac{2}{3}-\frac{3}{8}$
$\frac{8}{11}÷[\frac{2}{5}÷(1-\frac{7}{10})]$
$\frac{2}{3}÷\frac{5}{6}×\frac{2}{5}$
$\frac{5}{12}÷6+\frac{1}{6}×\frac{1}{3}$
答案
3. $8\frac{1}{8}$
$6$
$1$
$\frac{6}{11}$
$\frac{8}{25}$
$\frac{1}{8}$
$6$
$1$
$\frac{6}{11}$
$\frac{8}{25}$
$\frac{1}{8}$
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