Question

Am nevoie doar de B si de C. Multumesc

Answer 1

x ⚬ y = 2x + y - 3,  ∀ x,y ∈ ℤ

b)  x ⚬ (3x-1) = 6

⇔  2x + (3x-1) - 3 = 6

⇔  2x + 3x - 1 - 3 = 6

⇔  5x - 4 = 6

⇔  5x = 10

⇒  x = 2

c)  (x+1) ⚬ 3 = 5 ⚬ (x²-8)

⇔  2(x+1) + 3 - 3 = 2·5 + (x²-8) - 3

⇔  2x + 2 = 10 + x² - 8 - 3

⇔  2x + 2 = x² - 1

⇔  x² - 2x - 3 = 0

⇔  x² + x - 3x - 3 = 0

⇔  x(x+1) - 3(x+1) = 0

⇔  (x+1)(x-3) = 0

⇒  x ∈ {-1, 3}