`A = (1^2)/(1.2) . (2^2)/(2.3) . (3^2)/(3.4) . (4^2)/(4.5)`
` = (1^2. 2^2 . 3^2 . 4^2)/ (1.2.2.3.3.4.4.5)`
` = (1^2 .2^2 . 3^2 . 4^2)/ (1^2 . 2^2 . 3^2 . 4^2 . 5)`
` = 1/5`

`B = (2^2)/(1.3) . (3^2)/(2.4) . (4^2)/(3.5) . (5^2)/(4.6)`
`=  ( 2^2 . 3^2 . 4^2 . 5^2) / (1 . 3 . 2 . 4 . 3 . 5 . 4 .6)`
` = ( 2 . 2 . 3 . 3 . 4 . 4 . 5 . 5) / (1 . 3 . 2 . 4 . 3 . 5 . 4 .6)`
` = (( 2.3 .4 .5) . (2.3.4.5))/ ( 1. (2.3.4.5) . ( 3.4.6)`
` = ( 2.3.4.5) / (3.4.6)`
` = ( 2.3.4.5)/(3.4.2.3)`
` = 5/3`

Tính tổng :

`a) 2/(1.3) + 2/(3.5) + 2/(5.7) + ... +2/(99.101)`
` = 1 - 1/3 + 1/3- 1/5 + 1/5-1/7 +.... + 1/99-1/101`
` = 1 - 1/101`
` = 100/101`

`b) 5/(1.3) + 5/(3.5) + 5/(5.7) + ... + 5/ (99.101)`
` =5/2. [2/(1.3) + 2/(3.5) + 2/(5.7) + ... +2/(99.101)]`
` =5/2 .( 1 - 1/3 + 1/3- 1/5 + 1/5-1/7 +.... + 1/99-1/101)`
` = 5/2 . (1-1/101)`
` = 5/2 . 100/101`
` = 250/101`

 (-) A = $\dfrac{1^{2}}{1.2}$ . $\dfrac{2^{2}}{2.3}$ . $\dfrac{3^{2}}{3.4}$ .$\dfrac{4^{2}}{4.5}$ 

A = $\dfrac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}$ 

A = $\dfrac{(1.2.3.4).(1.2.3.4)}{(1.2.3.4).(2.3.4.5}$ 

A = $\dfrac{1}{5}$ ( rút gọn ở tử và mẫu ) 

B = $\dfrac{2^{2}}{1.3}$ . $\dfrac{3^{2}}{2.4}$ . $\dfrac{4^2}{3.5}$ . $\dfrac{5^{2}}{4.6}$ 

B= $\dfrac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}$ 

B = $\dfrac{(2.3.4.5).(2.3.4.5)}{(1.2.3.4).(3.4.5.6)}$ 

B = $\dfrac{5.2}{6}$ = $\dfrac{5}{3}$ 

(-) 

a. $\dfrac{2}{1.3}$ + $\dfrac{2}{3.5}$ + $\dfrac{2}{5.7}$ + .... + $\dfrac{2}{99.101}$ 

= 1-$\dfrac{1}{3}$ + $\dfrac{1}{3}$ - $\dfrac{1}{5}$ + $\dfrac{1}{5}$ - $\dfrac{1}{7}$ + .....+$\dfrac{1}{99}$ - $\dfrac{1}{101}$ 

= 1 - $\dfrac{1}{101}$ 

= $\dfrac{100}{101}$ 

b. $\dfrac{5}{1.3}$ + $\dfrac{5}{3.5}$ + $\dfrac{5}{5.7}$ + ... +$\dfrac{5}{99.101}$ 

= 5. $\dfrac{2}{2}$ . ( $\dfrac{1}{1.3}$ + $\dfrac{1}{3.5}$ + $\dfrac{1}{5.7}$ + ... +$\dfrac{1}{99.101}$ )

= $\dfrac{5}{2}$ . ( $\dfrac{2}{1.3}$ + $\dfrac{2}{3.5}$ + $\dfrac{2}{5.7}$ + .... + $\dfrac{2}{99.101}$ )

= $\dfrac{5}{2}$ . (1- $\dfrac{1}{3}$ + $\dfrac{1}{3}$ - $\dfrac{1}{5}$ + $\dfrac{1}{5}$ - $\dfrac{1}{7}$ + ....+$\dfrac{1}{99}$ - $\dfrac{1}{101}$ )

= $\dfrac{5}{2}$ . ( 1 - $\dfrac{1}{101}$ )

= $\dfrac{5}{2}$ . $\dfrac{100}{101}$ 

= $\dfrac{250}{101}$