Răspuns:
Explicație pas cu pas:
x o y = 4·(x+y-3) -xy
a) 2 o 4 = 4·(2+4-3) - 2·4 = 12-8 = 4
b) x o y = 4 - (x-4)(y-4) ?
x o y = 4x+4y-12-xy = x(4-y) +4(y-4) +4 =
= (y-4)(4-x) + 4 = 4 - (x-4)(y-4)
c) x o x o x o x = x
x o x o x o x = (x o x) o (x o x) = >
x o x = 4(x+x-3)-x·x = 4·(2x-3) -x² = -x² + 8x-12
(x o x) o (x o x) = (-x² + 8x-12) o (-x² + 8x-12) =
= 4(-x² + 8x-12-x² + 8x-12-3)- (-x² + 8x-12)² =
= 4(-2x²+16x-27) - (-x²+8x-12)² =
= -8x²+64x-108 - (x⁴-16x³+88x²-192x+144) =
= -8x²+64x-108 -x⁴+16x³-88x²+192x-144 =
= -x⁴+16x³-96x²+256x-252 =>
-x⁴+16x³-96x²+256x-252 = x =>
x⁴-16x³+96x²-255x+252 = 0
252 = 2²·3²·7
x⁴-16x³+96x²-255x+252 :(x-3) = x³-13x²+57x-84
x³-13x²+57x-84 :(x-4) = x²-9x+21
x²-9x+21 = 0 => x₃,₄ = (9±i√3)/2 ∉ R
Solutii x₁ = 3 ; x₂ = 4
