Răspuns:
E(x) = 4x²-6x+3 ; ∀ x ∈ R
n = {-1 ; 0 ; 1 ; 2}
Explicație pas cu pas:
E(x) = (x+4)²-2(2-x)(2x-3)-(x-3)²-2(3x+8)
E(x) = x²+8x+16+(2x-4)(2x-3)-(x²-6x+9)-6x-16
E(x) = x²+2x +4x²-6x-8x+12-x²+6x-9 =>
E(x) = 4x²-6x+3 ; ∀ x ∈ R
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E(n) ≤ -2n+11 <=>
4n²-6n+3 ≤ -2n+11 <=>
4n²-4n-8 ≤ 0 I:4 =>
n²-n-2 ≤ 0
n²-n-2 = 0 ; a = 1 ; b = -1 ; c = -2
Δ = b²-4ac = (-1)²-4·1·(-2) = 9 => √Δ = 3
n₁,₂ = (-b±√Δ)/2a = (1±3)/2 => n₁ = -1 ; n₂ = 2
n I -∞ -1 2 +∞
n²-n-2 I +++++++0-----0++++++ =>
n ∈ [-1 ; 2] I n ∈ Z => n = {-1 ; 0 ; 1 ; 2}
