Question

Mă poate ajuta cineva vă rog frumos?
Exercițiile 1,4​

Answer 1

Răspuns:

Explicație pas cu pas:

1)

2x²-5x-2 = 0 ; a = 2 ; b = -5 ; c = -2

2,5x²-3x = 0 ; a = 2,5 ; b = -3 ; c = 0

-x²+4x-2 = 0 ; a = -1 ; b = 4 ; c = -2

2x-x² = 0 ; a = -1 ; b = 2 ; c = 0

2-3x² = 0 ; a = -3 ; b = 0 ; c = 2

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a = 1 ; b = -1 ; c = 2 => x²-x+2 = 0

a = -4 ; b = 4 ; c = 0 => -4x²+4x = 0

a = √2 ; b = 0 ; c = -1 => √2x -1 = 0

a = -1 ; b = 2 ; c = -3 => -x²+2x-3 = 0

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4)a) {x∈R I 2x²+4x+2 = 0}

2x²+4x+2 = 0 <=> 2(x²+2x+1) = 0

<=> 2(x+1)² = 0 => x₁ = x₂ = -1 =>

=> multimea are un element

b)  {x∈R I 2x²-3x+1 = 0}

2x²-3x+1 = 0 <=> 2x²-2x-x+1 = 0 <=>

2x(x-1)-(x-1) = 0 <=> (2x-1)(x-1) = 0 =>

x₁ = 1/2 ; x₂ = 1 =>  multimea are doua elemente

c) {x∈R I 4x²-4x-1 = 0} ;

Δ = (-4)²-4·4·(-1) = 16+16 = 32

x₁,₂ = (4±√32)/8 = (4±4√2)/8 = (1±√2)/2

=>  multimea are doua elemente

d) {x∈R I x²+3x+2 = 0}

x²+3x+2 = 0 <=> x²+2x+x+2 = 0 <=>

x(x+2)+(x+2) = 0 <=> (x+1)(x+2) = 0 =>

x₁ = -2 ; x₂ = -1 =>

=>  multimea are doua elemente