Giải thích các bước giải:
Ta có:
$A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{37.38}$
$\to A=\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+\dfrac{38-37}{37.38}$
$\to A=\dfrac11-\dfrac12+\dfrac13-\dfrac14+...+\dfrac1{37}-\dfrac1{38}$
$\to A=(\dfrac11+\dfrac13+...+\dfrac1{37})-(\dfrac12+\dfrac14+...+\dfrac1{38})$
$\to A=(\dfrac11+\dfrac13+...+\dfrac1{37})+(\dfrac12+\dfrac14+...+\dfrac1{38})-2(\dfrac12+\dfrac14+...+\dfrac1{38})$
$\to A=\dfrac11+\dfrac12+...+\dfrac1{37}+\dfrac{1}{38}-(1+\dfrac12+...+\dfrac1{19})$
$\to A=\dfrac1{20}+\dfrac1{21}+...+\dfrac1{38}$
Ta có:
$B=\dfrac1{20.38}+\dfrac1{21.37}+...+\dfrac{1}{38.20}$
$\to B=\dfrac1{20.38}+\dfrac1{21.37}+...+\dfrac{1}{29.29}+...+\dfrac{1}{38.20}$
$\to B=\dfrac2{20.38}+\dfrac2{21.37}+...+\dfrac{2}{28.30}+\dfrac{1}{29.29}$
$\to B=2(\dfrac1{20.38}+\dfrac1{21.37}+...+\dfrac{1}{28.30})+\dfrac{1}{29.29}$
$\to 58B=2(\dfrac{58}{20.38}+\dfrac{58}{21.37}+...+\dfrac{58}{28.30})+\dfrac{58}{29.29}$
$\to 58B=2(\dfrac{20+38}{20.38}+\dfrac{21+37}{21.37}+...+\dfrac{28+30}{28.30})+\dfrac{29+29}{29.29}$
$\to 58B=2(\dfrac{1}{20}+\dfrac{1}{38}+\dfrac1{21}+\dfrac{1}{37}+..+\dfrac{1}{28}+\dfrac1{30})+\dfrac{2}{29}$
$\to 58B=2(\dfrac{1}{20}+\dfrac{1}{38}+\dfrac1{21}+\dfrac{1}{37}+..+\dfrac{1}{28}+\dfrac1{30}+\dfrac{1}{29})$
$\to 58B=2(\dfrac{1}{20}+\dfrac1{21}+...+\dfrac{1}{38})$
$\to 29B=\dfrac{1}{20}+\dfrac1{21}+...+\dfrac{1}{38}$
$\to 29B=A$
$\to \dfrac{A}{B}=29\in Z$
