Question

Rapid, ajutor va rog
$\sqrt{x ^{2} - 3 } = 1$
$\sqrt{2x + 3 } = x$
$\sqrt{2 + x} = x$
$\sqrt{x^{2} + 1} = 2$

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Answer 1

Răspuns:

Explicație pas cu pas:

√(x²-3) = 1   I²  =>

x²-3 = 1  => x² = 4 => x₁,₂ = ±2

Conditia radicalului : x²-3 ≥ 0 => x₁,₂ ∈ R - (-√3 ; +√3) =>

x₁ = -2 ; x₂ = 2 sunt solutii

√(2x+3) = x  I² ; conditii : 2x+3 ≥ 0 ; x ≥ 0 => x ∈ [0 ; +∞)

=> 2x+3 = x² => x²-2x-3 = 0 => x₁,₂ = [2±√(4+12)]/2

x₁,₂ = 1±2 ; x₁ = -1 ; x₂ = 3  => Solutie x = 3

√(2+x) = x  I²  => 2+x = x²  => x²-x-2 = 0 si

conditii: x ≥ -2 ; x ≥ 0 => x ∈ [0 ; +∞)

x₁,₂ = [1±√(1+8)]/2 = (1±3)/2 =>

x₁ = -1 ; x₂ = 2 => Solutie : x = 2

√(x²+1) = 2 I² ; x²+1 > 0 ; ∀x ∈ R =>

x²+1 = 4 => x² = 3 => x₁,₂ = ±√3

Solutii : x₁ = -√3 ; x₂ = √3