2025年8分钟暑假重点题五年级数学人教版
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$\frac {9}{10}
$\frac {11}{12}
-
\frac {2}{5}= \frac {1}{2}$ $\frac {3}{4}+
\frac {1}{5}= \frac {19}{20}$$\frac {11}{12}
+
\frac {1}{4}+
\frac {1}{12}= \frac {5}{4}$ $\frac {6}{7}-
\frac {13}{42}-
\frac {1}{6}= \frac {8}{21}$
答案:
- + + + - -
(1)$\frac {2}{3}+\frac {3}{5}= \frac {5}{8}$(
×
)$\frac {2}{3}+\frac {3}{5}=\frac {19}{15}$
(2)$\frac {5}{14}-\frac {1}{7}= \frac {4}{7}$(×
)$\frac {5}{14}-\frac {1}{7}=\frac {3}{14}$
答案:
(1) × $\frac {2}{3}+\frac {3}{5}=\frac {19}{15}$
(2) × $\frac {5}{14}-\frac {1}{7}=\frac {3}{14}$
(1) × $\frac {2}{3}+\frac {3}{5}=\frac {19}{15}$
(2) × $\frac {5}{14}-\frac {1}{7}=\frac {3}{14}$
3脱式计算。
$\frac {1}{8}+\frac {2}{15}+\frac {7}{8}$ $\frac {5}{6}+\frac {3}{4}-\frac {1}{3}$ $\frac {11}{12}-(\frac {1}{6}+\frac {1}{8})$
$11-\frac {7}{10}-\frac {3}{10}$ $\frac {7}{12}-(\frac {3}{4}-\frac {1}{2})$ $\frac {1}{2}-(\frac {3}{4}-\frac {3}{8})$
$\frac {5}{6}-\frac {1}{4}+\frac {1}{3}$ $\frac {1}{5}+\frac {7}{15}+\frac {4}{15}$ $\frac {7}{10}-(\frac {3}{4}-\frac {2}{5})$
$\frac {1}{8}+\frac {2}{15}+\frac {7}{8}$ $\frac {5}{6}+\frac {3}{4}-\frac {1}{3}$ $\frac {11}{12}-(\frac {1}{6}+\frac {1}{8})$
$11-\frac {7}{10}-\frac {3}{10}$ $\frac {7}{12}-(\frac {3}{4}-\frac {1}{2})$ $\frac {1}{2}-(\frac {3}{4}-\frac {3}{8})$
$\frac {5}{6}-\frac {1}{4}+\frac {1}{3}$ $\frac {1}{5}+\frac {7}{15}+\frac {4}{15}$ $\frac {7}{10}-(\frac {3}{4}-\frac {2}{5})$
答案:
$\frac {1}{8}+\frac {2}{15}+\frac {7}{8}$ $\frac {5}{6}+\frac {3}{4}-\frac {1}{3}$ $\frac {11}{12}-(\frac {1}{6}+\frac {1}{8})$
$=1\frac {2}{15}$ $=1\frac {1}{4}$ $=\frac {5}{8}$
$11-\frac {7}{10}-\frac {3}{10}$ $\frac {7}{12}-(\frac {3}{4}-\frac {1}{2})$ $\frac {1}{2}-(\frac {3}{4}-\frac {3}{8})$
$=10$ $=\frac {1}{3}$ $=\frac {1}{8}$
$\frac {5}{6}-\frac {1}{4}+\frac {1}{3}$ $\frac {1}{5}+\frac {7}{15}+\frac {4}{15}$ $\frac {7}{10}-(\frac {3}{4}-\frac {2}{5})$
$=\frac {11}{12}$ $=\frac {14}{15}$ $=\frac {7}{20}$
$=1\frac {2}{15}$ $=1\frac {1}{4}$ $=\frac {5}{8}$
$11-\frac {7}{10}-\frac {3}{10}$ $\frac {7}{12}-(\frac {3}{4}-\frac {1}{2})$ $\frac {1}{2}-(\frac {3}{4}-\frac {3}{8})$
$=10$ $=\frac {1}{3}$ $=\frac {1}{8}$
$\frac {5}{6}-\frac {1}{4}+\frac {1}{3}$ $\frac {1}{5}+\frac {7}{15}+\frac {4}{15}$ $\frac {7}{10}-(\frac {3}{4}-\frac {2}{5})$
$=\frac {11}{12}$ $=\frac {14}{15}$ $=\frac {7}{20}$
4疑难点解方程。
$x+\frac {1}{4}= \frac {2}{3}$ $x-\frac {4}{15}= \frac {1}{3}$ $\frac {5}{8}+x= \frac {11}{12}$ $\frac {19}{20}-x= \frac {1}{40}$
$x+\frac {1}{4}= \frac {2}{3}$ $x-\frac {4}{15}= \frac {1}{3}$ $\frac {5}{8}+x= \frac {11}{12}$ $\frac {19}{20}-x= \frac {1}{40}$
答案:
$x=\frac {5}{12}$ $x=\frac {3}{5}$ $x=\frac {7}{24}$ $x=\frac {37}{40}$
5素养点计算下面各题,你能发现什么规律?
$\frac {1}{2}+\frac {1}{3}=$
$\frac {1}{6}-\frac {1}{7}=$
我发现:当两个分数的分子都是1,分母只有公因数1时,它们相加(或减)的结果中,分母是两个分母的
$\frac {1}{2}+\frac {1}{3}=$
$\frac {5}{6}$
$\frac {1}{4}+\frac {1}{5}=$$\frac {9}{20}$
$\frac {1}{3}+\frac {1}{7}=$$\frac {10}{21}$
$\frac {1}{6}+\frac {1}{13}=$$\frac {19}{78}$
$\frac {1}{6}-\frac {1}{7}=$
$\frac {1}{42}$
$\frac {1}{5}-\frac {1}{8}=$$\frac {3}{40}$
$\frac {1}{8}-\frac {1}{9}=$$\frac {1}{72}$
$\frac {1}{9}-\frac {1}{10}=$$\frac {1}{90}$
我发现:当两个分数的分子都是1,分母只有公因数1时,它们相加(或减)的结果中,分母是两个分母的
积
,分子是两个分母的和
或差
。
答案:
$\frac {5}{6}$ $\frac {9}{20}$ $\frac {10}{21}$ $\frac {19}{78}$ $\frac {1}{42}$ $\frac {3}{40}$ $\frac {1}{72}$ $\frac {1}{90}$ 积
和 积 差
和 积 差
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