《2025年课堂笔记六年级数学上册人教版》

典例 用简便方法计算下面各题。
（1）$24× \frac{51}{43}+51× \frac{19}{43}$
（2）$\frac{7}{33}× \frac{8}{23}+\frac{8}{33}× \frac{16}{23}$

答案：

正确解答

（1）方法一：$24× \frac{51}{43}+51× \frac{19}{43}$

= $51× \frac{24}{43}+51× \frac{19}{43}$

= $51× (\frac{24}{43}+\frac{19}{43})$

= $51× 1$

=51

方法二：$24× \frac{51}{43}+51× \frac{19}{43}$

= $24× \frac{51}{43}+19× \frac{51}{43}$

= $\frac{51}{43}× (24 + 19)$

= $\frac{51}{43}× 43$

=51

（2）方法一：$\frac{7}{33}× \frac{8}{23}+\frac{8}{33}× \frac{16}{23}$

= $\frac{8}{33}× \frac{7}{23}+\frac{8}{33}× \frac{16}{23}$

= $\frac{8}{33}× (\frac{7}{23}+\frac{16}{23})$

= $\frac{8}{33}× 1$

=$\frac{8}{33}$

方法二：$\frac{7}{33}× \frac{8}{23}+\frac{8}{33}× \frac{16}{23}$

= $\frac{7}{33}× \frac{8}{23}+\frac{16}{33}× \frac{8}{23}$

= $\frac{8}{23}× (\frac{7}{33}+\frac{16}{33})$

= $\frac{8}{23}× \frac{23}{33}$

=$\frac{8}{33}$