Răspuns:
x₁ = 1 ; x₂ = 1/2 ; x₃ = 3/2
Explicație pas cu pas:
x ο y = 2xy - 2x -2y +3
a) 2020 ο 1 = 1 ??
2020 ο 1 = 2·2020·1 - 2·2020 - 2·1 + 3 =
= 4040 - 4040 - 2 +3 = 3-2 = 1
b) x ο y = 2·(x-1)·(y-1)+1 ??
x ο y = 2xy - 2x -2y +3 = 2x·(y-1) - 2y + 2 +1 =
= 2x·(y-1) - 2·(y-1) + 1 = (y-1)·(2x-2) + 1 = 2·(x-1)·(y-1) + 1
c) (x ο x) ο x = x => x = ??
x ο x = 2x·x - 2x - 2x + 3 = 2x² - 4x + 3
(x ο x) ο x = (2x²-4x+3) ο x =
= 2·(2x²-4x+3)·x - 2·(2x²-4x+3) - 2x + 3 =
= (4x²-8x+6)·x - (4x²-8x+6) - 2x +3 =
= 4x³-8x²+6x - 4x²+8x-6-2x+3 =
= 4x³-12x²+12x-3 =>
4x³-12x²+12x-3 = x <=>
4x³-12x²+11x-3 = 0
x₁ = 1 (4·1³-12·1²+11·1-3 = 4-12+11-3 = 15-15=0)
4x³-12x²+11x-3 : (x-1) = 4x²-8x+3
-4x³+4x²
_______
/ -8x²+11x-3
8x²-8x
________
/ 3x-3
-3x+3
______
/ /
4x²-8x+3 = 0 ; a = 4 ; b = -8 ; c = 3
Δ = b²-4ac = (-8)²-4·4·3 = 64 - 48 = 16
√Δ = √16 = 4
=> x₂,₃ = (-b ±√Δ)/2a = (8±4)/8 =>
x₂ = (8-4)/8 = 4/8 = 1/2
x₃ = (8+4)/8 = 12/8 = 3/2
Solutii : x₁ = 1 ; x₂ = 1/2 ; x₃ = 3/2
