《2025年实验班提优训练暑假衔接版五升六年级数学北师大版》

例 计算:$(\frac {1}{11}+\frac {1}{21}+\frac {1}{31}+\frac {1}{41})×(\frac {1}{21}+\frac {1}{31}+\frac {1}{41}+\frac {1}{51})-(\frac {1}{11}+\frac {1}{21}+\frac {1}{31}+\frac {1}{41}+\frac {1}{51})×(\frac {1}{21}+\frac {1}{31}+\frac {1}{41})$
思路点拨 把算式中相同的一部分用字母代替,可以化繁为简,这种方法叫作代数法。当算式中有相同的部分,可以用代数法解答。
答案详解 设$A= \frac {1}{21}+\frac {1}{31}+\frac {1}{41},B= \frac {1}{11}+\frac {1}{21}+\frac {1}{31}+\frac {1}{41}$。
$(\frac {1}{11}+\frac {1}{21}+\frac {1}{31}+\frac {1}{41})×(\frac {1}{21}+\frac {1}{31}+\frac {1}{41}+\frac {1}{51})-(\frac {1}{11}+\frac {1}{21}+\frac {1}{31}+\frac {1}{41}+\frac {1}{51})×(\frac {1}{21}+\frac {1}{31}+\frac {1}{41})$
$=B×(A+\frac {1}{51})-(B+\frac {1}{51})×A$
$=AB+\frac {1}{51}B-AB-\frac {1}{51}A$
$=\frac {1}{51}(B-A)$
$=\frac {1}{51}×\frac {1}{11}$
$=\frac {1}{561}$

答案： $\frac {1}{561}$

3. 计算:$(\frac {1}{8}+\frac {1}{9}+\frac {1}{10}+\frac {1}{11})×(\frac {1}{9}+\frac {1}{10}+\frac {1}{11}+\frac {1}{12})-(\frac {1}{8}+\frac {1}{9}+\frac {1}{10}+\frac {1}{11}+\frac {1}{12})×(\frac {1}{9}+\frac {1}{10}+\frac {1}{11})$

答案： $\frac{1}{96}$

4. 计算:$(1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5})×(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6})-(1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6})×(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+$$\frac {1}{5})$

答案： $\frac{1}{6}$