例2 先觀察,再通過(guò)計(jì)算比較大小。
$ \frac{4}{1 × 3} ? \frac{1}{1} + \frac{1}{3} $
$ \frac{8}{3 × 5} ? \frac{1}{3} + \frac{1}{5} $
$ \frac{12}{5 × 7} ? \frac{1}{5} + \frac{1}{7} $
$ \frac{16}{7 × 9} ? \frac{1}{7} + \frac{1}{9} $
(1) 根據(jù)上面算式中蘊(yùn)含的規(guī)律再寫(xiě)一道這樣的算式：
$\frac{28}{13×15}=\frac{1}{13}+\frac{1}{15}$（答案不唯一）
；
(2) 計(jì)算：$ \frac{4}{1 × 3} - \frac{8}{3 × 5} + \frac{12}{5 × 7} - \frac{16}{7 × 9} + \frac{20}{9 × 11} - \frac{24}{11 × 13} 。$
分析：通過(guò)計(jì)算可知,每組中兩道算式的結(jié)果相等。(1) 通過(guò)觀察可以發(fā)現(xiàn)上面算式的特點(diǎn)：分?jǐn)?shù)的分母是兩個(gè)數(shù)相乘的積,分子是這兩個(gè)數(shù)相加的和,這樣的分?jǐn)?shù)值的大小等于分子是1,分母分別是這兩個(gè)數(shù)的兩個(gè)分?jǐn)?shù)之和。符合這一特點(diǎn)的算式有無(wú)數(shù)道,如：$ \frac{19}{8 × 11} = \frac{1}{8} + \frac{1}{11} ；$(2) 根據(jù)(1)中發(fā)現(xiàn)的規(guī)律,我們可以把算式中的每個(gè)分?jǐn)?shù)看作兩個(gè)分?jǐn)?shù)的和,再通過(guò)去括號(hào)將一些分?jǐn)?shù)互相抵消,從而使計(jì)算簡(jiǎn)便。
解答：= = = =
(2) $\begin{aligned} & \frac{4}{1 × 3} - \frac{8}{3 × 5} + \frac{12}{5 × 7} - \frac{16}{7 × 9} + \frac{20}{9 × 11} - \frac{24}{11 × 13} \\ = & \left( \frac{1}{1} + \frac{1}{3} \right) - \left( \frac{1}{3} + \frac{1}{5} \right) + \left( \frac{1}{5} + \frac{1}{7} \right) - \left( \frac{1}{7} + \frac{1}{9} \right) + \left( \frac{1}{9} + \frac{1}{11} \right) - \left( \frac{1}{11} + \frac{1}{13} \right) \\ = & \frac{1}{1} + \frac{1}{3} - \frac{1}{3} - \frac{1}{5} + \frac{1}{5} + \frac{1}{7} - \frac{1}{7} - \frac{1}{9} + \frac{1}{9} + \frac{1}{11} - \frac{1}{11} - \frac{1}{13} \\ = & 1 - \frac{1}{13} \\ = & \frac{12}{13} \end{aligned}$
答案：(1) $\frac{28}{13×15}=\frac{1}{13}+\frac{1}{15}$（答案不唯一）
(2) $\begin{aligned} & \frac{4}{1 × 3} - \frac{8}{3 × 5} + \frac{12}{5 × 7} - \frac{16}{7 × 9} + \frac{20}{9 × 11} - \frac{24}{11 × 13} \\ = & \left( \frac{1}{1} + \frac{1}{3} \right) - \left( \frac{1}{3} + \frac{1}{5} \right) + \left( \frac{1}{5} + \frac{1}{7} \right) - \left( \frac{1}{7} + \frac{1}{9} \right) + \left( \frac{1}{9} + \frac{1}{11} \right) - \left( \frac{1}{11} + \frac{1}{13} \right) \\ = & \frac{1}{1} + \frac{1}{3} - \frac{1}{3} - \frac{1}{5} + \frac{1}{5} + \frac{1}{7} - \frac{1}{7} - \frac{1}{9} + \frac{1}{9} + \frac{1}{11} - \frac{1}{11} - \frac{1}{13} \\ = & 1 - \frac{1}{13} \\ = & \frac{12}{13} \end{aligned}$
$5. $計(jì)算：$ 1 \frac{1}{2} - \frac{5}{6} + \frac{7}{12} - \frac{9}{20} + \frac{11}{30} - \frac{13}{42} + \frac{15}{56} 。$
答案：$1\frac{1}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}$
$=(1+\frac{1}{2})-(\frac{1}{2}+\frac{1}{3})+(\frac{1}{3}+\frac{1}{4})-(\frac{1}{4}+\frac{1}{5})+(\frac{1}{5}+\frac{1}{6})-(\frac{1}{6}+\frac{1}{7})+(\frac{1}{7}+\frac{1}{8})$
$=1+\frac{1}{2}-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}-\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}=1+\frac{1}{8}=\frac{9}{8}$
提示：把算式中的每個(gè)分?jǐn)?shù)看作兩個(gè)分?jǐn)?shù)的和，再通過(guò)去括號(hào)等將一些分?jǐn)?shù)互相抵消，從而使計(jì)算簡(jiǎn)便。
$6. $計(jì)算：$ \frac{49}{12} - \frac{63}{20} + \frac{77}{30} - \frac{91}{42} + \frac{105}{56} 。$
答案：$\frac{49}{12}-\frac{63}{20}+\frac{77}{30}-\frac{91}{42}+\frac{105}{56}$
$=7×(\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56})$
$=7×[(\frac{1}{3}+\frac{1}{4})-(\frac{1}{4}+\frac{1}{5})+(\frac{1}{5}+\frac{1}{6})-(\frac{1}{6}+\frac{1}{7})+(\frac{1}{7}+\frac{1}{8})]$
$=7×(\frac{1}{3}+\frac{1}{4}-\frac{1}{4}-\frac{1}{5}+\frac{1}{5}+\frac{1}{6}-\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8})$
$=7×(\frac{1}{3}+\frac{1}{8})=7×\frac{11}{24}=\frac{77}{24}$
提示：提取公因數(shù)7后的算式的每個(gè)分?jǐn)?shù)可以看作兩個(gè)分?jǐn)?shù)的和，再通過(guò)去括號(hào)等將一些分?jǐn)?shù)互相抵消，從而使計(jì)算簡(jiǎn)便。
$7. $計(jì)算：$ \frac{20}{6} - \frac{28}{12} + \frac{36}{20} - \frac{44}{30} + \frac{52}{42} - \frac{60}{56} 。$
答案：$\frac{20}{6}-\frac{28}{12}+\frac{36}{20}-\frac{44}{30}+\frac{52}{42}-\frac{60}{56}$
$=(\frac{5}{6}-\frac{7}{12}+\frac{9}{20}-\frac{11}{30}+\frac{13}{42}-\frac{15}{56})×4$
$=(\frac{1}{2}+\frac{1}{3}-\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+\frac{1}{5}-\frac{1}{5}-\frac{1}{6}+\frac{1}{6}+\frac{1}{7}-\frac{1}{7}-\frac{1}{8})×4=(\frac{1}{2}-\frac{1}{8})×4=\frac{3}{8}×4=\frac{3}{2}$
提示：提取公因數(shù)4后的算式的每個(gè)分?jǐn)?shù)可以看作兩個(gè)分?jǐn)?shù)的和，再通過(guò)去括號(hào)等將一些分?jǐn)?shù)互相抵消，從而使計(jì)算簡(jiǎn)便。
$8. $計(jì)算：$ \frac{4}{1 × 3} - \frac{8}{3 × 5} + \frac{12}{5 × 7} - \frac{16}{7 × 9} + … + \frac{4052}{2025 × 2027} - \frac{4056}{2027 × 2029} 。$
答案：$\frac{4}{1×3}-\frac{8}{3×5}+\frac{12}{5×7}-\frac{16}{7×9}+\cdots+\frac{4052}{2025×2027}-\frac{4056}{2027×2029}$
$=(1+\frac{1}{3})-(\frac{1}{3}+\frac{1}{5})+(\frac{1}{5}+\frac{1}{7})-(\frac{1}{7}+\frac{1}{9})+\cdots+(\frac{1}{2025}+\frac{1}{2027})-(\frac{1}{2027}+\frac{1}{2029})$
$=1-\frac{1}{2029}$
$=\frac{2028}{2029}$
提示：利用裂項(xiàng)的方法，把每個(gè)分?jǐn)?shù)拆分成兩個(gè)分?jǐn)?shù)的和，利用減法的性質(zhì)去掉括號(hào)后，即可抵消，最后剩下1與最后一個(gè)分?jǐn)?shù)的差。
