Question

Exercitiul 4 va rog frumos.

Answer 1

6/(2x+1) ∈ Z ⇔  (2x+1) ∈D6      2x+1 ∈ {-6,-3,-2,-1,1,2,3,6} 
2x ∈ { -7,-4,-3,-2,0,1,2,5}    x∈ {-2,-1,0,1}     A = {-2,-1,0,1}
(2x +√3)(2-x√3) = 1  
4x +2√3 - 2√3 ·x² - 3x = 1
2√3·x² - x + 1 - 2√3 =0
Δ = 1 +8√3(2√3 - 1) = 1 +48 - 8√3 = 49 - 8√3  = (4√3 -1)²
x1 = (-1 + 4√3 -1)/4√3 = -2/4√3 + 1 = 1 - √3 /6
x2 = (-1 - 4√3 +1)/4√3 = - 1    B = {-1}
a) A∩B = -1
b)  S = - 2