`1) a. A = 1/2^2 + 1/3^2 + ... + 1/2012^2 + 1/2013^2`
`< 1/1.2 + 1/2.3 + ... + 1/2011.2012 + 1/2012.2013`
`= 1 - 1/2 + 1/2 - 1/3 + ...+ 1/2011 - 1/2012 + 1/2012 - 1/2013`
`= 1 - 1/2013 < 1`
` Vậy ` `A < 1 (đpcm)`
`b. A = 2/3 + 2/3^2 + 2/3^3 +...+ 2/3^2018`
`⇒ 1/2 A = 1/3 + 1/3^2 + 1/3^3 +...+ 1/3^2018`
`⇒ 1/6 A = 1/3^2 + 1/3^3 + 1/3^4 +...+ 1/3^2019`
`⇒ 1/2 A - 1/6 A = (1/3 + 1/3^2 + 1/3^3 +...+ 1/3^2018) - (1/3^2 + 1/3^3 + 1/3^4 +...+ 1/3^2019)`
`⇒ 1/3 A = 1/3 - 1/3^2019`
`⇒ A = (1/3 - 1/3^2019) . 3`
`⇒ A = 1 - 1/3^2018 < 1`
`2) 1/3 + 1/6 + 1/10 +....+ 2/(x.(x+1)) = 2015/2017`
`⇒ 2 . (1/6 + 1/12 + 1/20 +....+ 1/(x.(x+1))) = 2015/2017`
`⇒ 1/2.3 + 1/3.4 + 1/4.5 +....+ 1/(x.(x+1)) = 2015/2017 : 2``
`⇒ 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +... + 1/x - 1/(x+1) = 2015/4034`
`⇒ 1/2 - 1/(x+1) = 2015/4034`
`⇒ 1/(x+1) = 1/2 - 2015/4034`
`⇒ 1/(x+1) = 1/2017`
`⇒ x + 1 = 2017`
`⇒ x = 2016`