Question

16．遞等式計算，能簡算就簡算
$\frac{7}{15}$×$\frac{21}{37}$×$\frac{15}{7}$           
$\frac{9}{41}$×$\frac{23}{30}$+$\frac{9}{41}$×$\frac{7}{30}$            
$\frac{6}{13}$÷$\frac{6}{26}$÷$\frac{1}{4}$．

Answer 1

分析 ①$\frac{7}{15}$×$\frac{21}{37}$×$\frac{15}{7}$，運用乘法交換律進(jìn)行簡算；         
②$\frac{9}{41}$×$\frac{23}{30}$+$\frac{9}{41}$×$\frac{7}{30}$，運用乘法分配律簡算；          
③$\frac{6}{13}$÷$\frac{6}{26}$÷$\frac{1}{4}$，把除數(shù)轉(zhuǎn)化為乘它的倒數(shù)，再根據(jù)分?jǐn)?shù)乘法的計算法則計算．

解答 解：①$\frac{7}{15}$×$\frac{21}{37}$×$\frac{15}{7}$ 
=$\frac{7}{15}×\frac{15}{7}×\frac{21}{37}$
=$\frac{21}{37}$；
          
②$\frac{9}{41}$×$\frac{23}{30}$+$\frac{9}{41}$×$\frac{7}{30}$    
=$\frac{9}{41}×（\frac{23}{30}+\frac{7}{30}）$
=$\frac{9}{41}×1$
=$\frac{9}{41}$；
        
③$\frac{6}{13}$÷$\frac{6}{26}$÷$\frac{1}{4}$
=$\frac{6}{13}×\frac{26}{6}×4$
=8．

點評 此題考查的目的是理解掌握分?jǐn)?shù)四則混合運算的順序以及它們的計算法則，并且能夠靈活選擇簡便方法進(jìn)行計算．