Răspuns:
Explicație pas cu pas:
a) (-2x-12)·(x-7) ≥ 0, x ∈ Z <=>
-2x²+14x-12x+84 ≥ 0 <=>
-2x²+2x+84 ≥ 0 I:(-2) =>
x²-x-42 ≤ 0
x²-x-42 = 0 =>
a = 1 ; b = -1 ; c = -42
Δ = b²-4ac = 1+168 = 169
√Δ = √169 = 13
x₁,₂ = (-b±√Δ)/2a = (1±13)/2
x₁ = -6 ; x₂ = 7
x I -∞ -6 7 +∞
x²-x-42 I+++++++++0-------0++++++++
x ∈ [-6 ; 7] ; x ∈ Z =>
x ∈ {-6 ; -5 ; -4 ; -3 ; -2 ; -1 ; 0 ; 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7}
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b) (2x-12)·(2x+1) ≤ 0, x ∈ Z =>
4x²+2x-24x-12 ≤ 0 <=>
4x²-22x-12 ≤ 0 I:2 =>
2x²-11x-6 ≤ 0
2x²-11x-6 = 0 =>
a = 2 ; b = -11 ; c = -6
Δ = b²-4ac = 121+48 = 169 =>
√Δ = √169 = 13
x₁,₂ = (-b±√Δ)/2a = (11±13)/4
x₁ = -1/2 ; x₂ = 6
x I-∞ -1/2 6 +∞
2x²-11x-6 I +++++++0----------0+++++++
x ∈ [-1/2 ; 6] ; x ∈ Z =>
x ∈ {0 ; 1 ; 2 ; 3 ; 4 ; 5 ; 6}
