\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Vì 11 ∈ B(2n - 1) => 2n - 1 ∈ B(11)
=> B(11) = { + 1 ; + 11 }
Ta có : 2n - 1 = 1 <=> 2n = 2 => n = 1 ( TM )
2n - 1 = - 1 <=> 2n = 0 => n = 0 ( TM )
2n - 1 = 11 <=> 2n = 12 => n = 6 ( TM )
2n - 1 = - 11 <=> 2n = - 10 => n = - 5 ( TM )
Vậy n = { - 5; 0; 1; 6 }