《2025年小学升初中衔接读本数学》

例1 计算下列各题。
(1)$1260÷28+72×52$
(2)$0.8-0.8×0.56÷0.7$
(3)$[\frac {9}{10}-(\frac {1}{4}+\frac {7}{20})]÷\frac {4}{5}$
【解答】(1)$1260÷28+72×52$
$=45+3744$
$=3789$
(2)$0.8-0.8×0.56÷0.7$
$=0.8-0.448÷0.7$
$=0.8-0.64$
$=0.16$
(3)$[\frac {9}{10}-(\frac {1}{4}+\frac {7}{20})]÷\frac {4}{5}$
$=(\frac {9}{10}-\frac {3}{5})÷\frac {4}{5}$
$=\frac {3}{10}×\frac {5}{4}$
$=\frac {3}{8}$

答案：
(1)$1260÷28+72×52$
$=45+3744$
$=3789$
(2)$0.8-0.8×0.56÷0.7$
$=0.8-0.448÷0.7$
$=0.8-0.64$
$=0.16$
(3)$[\frac {9}{10}-(\frac {1}{4}+\frac {7}{20})]÷\frac {4}{5}$
$=(\frac {9}{10}-\frac {3}{5})÷\frac {4}{5}$
$=\frac {3}{10}×\frac {5}{4}$
$=\frac {3}{8}$

例2 用简便方法计算下列各题。
(1)$\frac {5}{9}+\frac {7}{9}-\frac {5}{8}-\frac {3}{8}$
(2)$1000÷125÷25÷8÷4$
(3)$51\frac {34}{35}÷17$
(4)$\frac {2}{3}×4+\frac {2}{3}-\frac {2}{3}×5$
【解答】(1)$\frac {5}{9}+\frac {7}{9}-\frac {5}{8}-\frac {3}{8}$
$=(\frac {5}{9}+\frac {7}{9})-(\frac {5}{8}+\frac {3}{8})$
$=\frac {4}{3}-1$
$=\frac {1}{3}$
(2)$1000÷125÷25÷8÷4$
$=1000÷(125×25×8×4)$
$=1000÷[(125×8)×(25×4)]$
$=0.01$
(3)$51\frac {34}{35}÷17$
$=(51+\frac {34}{35})×\frac {1}{17}$
$=51×\frac {1}{17}+\frac {34}{35}×\frac {1}{17}$
$=3\frac {2}{35}$
(4)$\frac {2}{3}×4+\frac {2}{3}-\frac {2}{3}×5$
$=\frac {2}{3}×(4+1-5)$
$=\frac {2}{3}×0$
$=0$

答案：
(1)$\frac {5}{9}+\frac {7}{9}-\frac {5}{8}-\frac {3}{8}$
$=(\frac {5}{9}+\frac {7}{9})-(\frac {5}{8}+\frac {3}{8})$
$=\frac {12}{9}-1$
$=\frac {4}{3}-1$
$=\frac {1}{3}$
(2)$1000÷125÷25÷8÷4$
$=1000÷(125×8×25×4)$
$=1000÷[(125×8)×(25×4)]$
$=1000÷(1000×100)$
$=1000÷100000$
$=0.01$
(3)$51\frac {34}{35}÷17$
$=(51+\frac {34}{35})×\frac {1}{17}$
$=51×\frac {1}{17}+\frac {34}{35}×\frac {1}{17}$
$=3+\frac {2}{35}$
$=3\frac {2}{35}$
(4)$\frac {2}{3}×4+\frac {2}{3}-\frac {2}{3}×5$
$=\frac {2}{3}×(4+1-5)$
$=\frac {2}{3}×0$
$=0$