`text{Ta có}`
`lim (x_(n+1))/(x_n)`
`= lim (\sqrt{a.x_(n)^2 + 3x_n + 2018})/(x_n)`
`= lim (\sqrt{a + 3/(x_n) + 2018/(x_(n)^2)})/1`
`= I`
`text{Ta có}`
`a > 1, x_(n+1) = sqrt{a.x_(n)^2 + 3x_n + 2018}`
`-> x_(n+1) > x_n`
`-> lim (3/(x_n)) = lim (2018/(x_(n)^2)) = 0`
`-> I = sqrt{a}`
`text{Giả thiết}`
`-> sqrt(a) = 2018`
`-> a = 2018^2`