Ta có : 

$A$= $\dfrac{1}{1.2}$ + $\dfrac{1}{3.4}$ + ... + $\dfrac{1}{2005.2006}$ 

$A$ = $\dfrac{1}{1}$ - $\dfrac{1}{2}$ + $\dfrac{1}{3}$ - $\dfrac{1}{4}$ + ... + $\dfrac{1}{2005}$ - $\dfrac{1}{2006}$ 

$A$ = ( $\dfrac{1}{1}$ + $\dfrac{1}{3}$ + $\dfrac{1}{5}$ + .... + $\dfrac{1}{2005}$  ) - ( $\dfrac{1}{2}$ + $\dfrac{1}{4}$ + ... + $\dfrac{1}{2006}$ )

$A$ = ( $\dfrac{1}{1}$ + $\dfrac{1}{2}$ + $\dfrac{1}{3}$ + $\dfrac{1}{4}$  + $\dfrac{1}{5}$ + $\dfrac{1}{6}$ + .... + $\dfrac{1}{2005}$ + $\dfrac{1}{2006}$ ) - $2$ . (  $\dfrac{1}{2}$ + $\dfrac{1}{4}$ + ... + $\dfrac{1}{2006}$ )

$A$ = ( $\dfrac{1}{1}$ + $\dfrac{1}{2}$ + $\dfrac{1}{3}$ + $\dfrac{1}{4}$  + $\dfrac{1}{5}$ + $\dfrac{1}{6}$ + .... + $\dfrac{1}{2005}$ + $\dfrac{1}{2006}$ ) - ( $1$ + $\dfrac{1}{2}$ + ... + $\dfrac{1}{1003}$ )

$A$ = $\dfrac{1}{1004}$ + $\dfrac{1}{1005}$ + ... + $\dfrac{1}{2006}$ ($1$)

Ta lại có : 

$B$ = $\dfrac{1}{1004 . 2006}$ + $\dfrac{1}{1005 . 2005 }$( cái này bị sai đề bạn nha ) + $...$ + $\dfrac{1}{2006 . 1004}$ 

$3010B$ = $\dfrac{3010}{1004 . 2006}$ + $\dfrac{3010}{1005 . 2005 }$ + .... + $\dfrac{3010}{2006 . 1004}$

$3010B$ = $\dfrac{1}{1004}$ + $\dfrac{1}{2006}$ + $\dfrac{1}{1005}$ + $\dfrac{1}{2005}$ + ... + $\dfrac{1}{2006}$ + $\dfrac{1}{1004}$ = $2$ . ( $\dfrac{1}{1004}$ + $\dfrac{1}{1005}$ + ... + $\dfrac{1}{2006}$ )

⇒ $B$ = $\dfrac{2 . ( \dfrac{1}{1004} + \dfrac{1}{1005} + ... + \dfrac{1}{2006} ) }{3010}$ = $\dfrac{\dfrac{1}{1004} + \dfrac{1}{1005} + ... + \dfrac{1}{2006} )}{1505}$ ($2$)

Từ ($1$) , ($2$) ⇒

$\dfrac{A}{B}$ = $\dfrac{\dfrac{1}{1004} + \dfrac{1}{1005} + ... + \dfrac{1}{2006}}{\dfrac{\dfrac{1}{1004} + \dfrac{1}{1005} + ... + \dfrac{1}{2006} )}{1505}}$ = $1505$ ∈ $Z$

⇒ $\dfrac{A}{B}$ ∈ $Z$ ( đpcm )

Giải thích các bước giải:

A=$\frac{1}{1.2}$ +$\frac{1}{3.4}$ +....+$\frac{1}{2005.2006}$ 

A=1-$\frac{1}{2}$ +$\frac{1}{3}$ -$\frac{1}{4}$ +....+$\frac{1}{2005}$ -$\frac{1}{2006}$ 

A=(1+$\frac{1}{3}$ +$\frac{1}{5}$ +...+$\frac{1}{2005}$ )-($\frac{1}{2}$ +$\frac{1}{4}$ +$\frac{1}{6}$ +.....+$\frac{1}{2006}$ )

A=(1+$\frac{1}{2}$ +$\frac{1}{3}$ +...+$\frac{1}{2006}$ )-2.($\frac{1}{2}$ +$\frac{1}{4}$ +$\frac{1}{6}$ +.....+$\frac{1}{2006}$ )

A=(1+$\frac{1}{2}$ +$\frac{1}{3}$ +...+$\frac{1}{2006}$ )-(1+$\frac{1}{2}$ +$\frac{1}{3}$ .....+$\frac{1}{1003}$ )

A=$\frac{1}{1004}$ +$\frac{1}{1005}$ +....+$\frac{1}{2006}$ 

B=$\frac{1}{1004}$ .$\frac{1}{2006}$ +$\frac{1}{1005}$ .2006+.....+$\frac{1}{1006}$ .1004 3010

B=1004+$\frac{2006}{1004}$ .2006+1005+$\frac{2006}{1005}$ .2006+.....+2006+$\frac{1004}{2006}$ .1004 3010

B=$\frac{1}{2006}$ +$\frac{1}{1004}$ +$\frac{1}{1005}$ +$\frac{1}{1005}$ +....+$\frac{1}{1004}$ +$\frac{1}{2006}$ 3010

B=2.(1/1004+1/1005+1/1006+....+1/2006)

B=2.(1/1004+1/1005+1/1006+....+1/2006)/3010

B=(1/1004+1/1005+1/1006+....+1/2006)/1505 ⇒A/

B=(1/1004+1/1005+....+1/2006)/(1/1004+1/1005+1/1006+....+1/2006)/1505 ⇒A/B=1505

Hay A/B ∈ Z

( Mk vt P/s thế thôi nhá còn đâu bn tự viết vào giúp mk. Bh mk bận nếu bn k hiểu thì tý mk vt tiếp cho)

$ Xin Ctlhn$

$@ nhuquynh1110$