найдите область определения D(y)
1) y = 5/(3x^2 - 2) 3x^2 - 2 ≠ 0 3x^2 ≠ 2 x^2 = 2/3 x1 = -√(2/3) x2 = √(2/3) D(y) = (- ≈ ; -√(2/3)) ( √(2/3) ; + ≈) 2) y = √(3 - x) 3 - x ≥ 0 x ≤ 3 D(y) = (- ≈ ; 3]