Question

Va rog ajutati-ma si pe mine la mate ​

Answer 1

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