Răspuns:
Explicație pas cu pas:
a) √(3x-12) = 2 I² => 3x-12 = 4 => 3x = 12+4 => x = 16/3
3x-12 ≥ 0 => 3x ≥ 12 => x ≥ 4 ; 16/3 ∈ I
b) √(2x-1) = 7 I² => 2x-1 = 49 => 2x = 50 => x = 25
2x-1 ≥ 0 => 2x ≥ 1 => x ≥ 1/2 ; 25 ∈ I
c) √(3x+4) = 2√x I² => 3x+4 = 4x => x = 4
3x+4 ≥ 0 => x ≥ -4/3}
x ≥ 0 } => x ≥ 0 ; 4 ∈ I
d) √(x+6) = x I² => x+6 = x² => x²-x-6 = 0 =>
x₁,₂ = [1±√(1+24)]/2 = (-1±5)/2 => x₁ = -3 ; x₂ = 2
x+6 ≥ 0 ; x ≥ 0 => x ≥ 0
Solutie : doar x = 2
e) √(x²-x-2) = x-2 I² => x²-x-2 = x²-4x+4
3x = 6 => Solutie : x = 2 ∈ I
x²-x-2 ≥ 0 ; x-2 ≥ 0
x₁,₂ = [1±√(1+8)]/2 = [1±3]/2 => x₁ = -1 ; x₂ = 2 =>
x ≥ 2 (din conditia radicalului)
f) √(x²+2x+6) = 3 I² => x²+2x+6 = 9 =>
x²+2x-3 = 0 => x₁,₂ = [-2±√(4+12)]/2 = [-2±4]/2
x₁ = -3 ; x₂ = 1
Conditia este x₂+2x+6 ≥ 0 ; adevarat pt. (∀)x ∈ R
g) √(x-1) = 3-x I² => x-1 = 9-6x+x²
Conditii : x-1 ≥0 => x ≥1 ; 3-x ≥ 0 => x ≤3 =>
x ∈ [1 , 3]
x-1 = 9-6x+x² => x²-7x+10 =0 =>
x₁,₂ = [7±√(49-40)]/2 = [7±3]/2 =>
x₁ = 2 ; x₂ = 5 => Solutie : doar x = 2
