Question

Legi de compozitie, am nevoie de ajutor, multumesc

Answer 1

Răspuns:

Explicație pas cu pas:

x o y = xy - 3x - 3y + 12

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3 o 4 - 2 o 3 = ?

3 o 4 = 3·4-3·3-3·4+12 = 12-9-12+12 = 3

2 o 3 = 2·3-3·2-3·3+12 = 6-6-9+12 = 3

3 o 4 - 2 o 3 = 3 - 3 =  0

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x o 4 = 5  <=>

x·4 - 3·x - 3·4 + 12 = 5 <=>

4x-3x - 12 + 12 = 5 =>

x = 5

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x o x = 4 =>

x·x - 3·x - 3·x + 12 = 4 =>

x²-6x + 12-4 = 0 =>

x²-6x+8 = 0 =>

a = 1 ; b = -6 ; c = 8 ; Δ = b²-4ac = 36-32 = 4 ; √Δ = √4 = 2

x₁,₂ = (-b±√Δ)/2a = (6±2)/2 = >

x₁ = (6-2)/2 = 4/2 => x₁ = 2

x₂ = (6+2)/2 = 8/2 => x₂ = 4

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x o 3 = 3 ? =>

x o 3 = x·3-3·x - 3·3 + 12 = 12-9 = 3 =>

x o 3 = 3 , (∀) x ∈ R

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A = 1 o 2 o 3 o ......o 20

1 o 2 = 1·2-3·1-3·2+12 = 2-3-6+12 = 14-9 = 5

1 o 2 o 3 = 5 o 3 = 5·3-3·5-3·3+12 = 15-15-9+12 = 3

1 o 2 o 3 o 4 = 3 o 4 = 3

1 o 2 o 3 o 4 o 5 = 3 o 5 = 3·5-3·3-3·5+12 = 12-9 = 3

1 o 2 o 3 o 4 o 5 o 6 = 3 o 6 = 3·6-3·6-3·3+12 = 12-9 = 3 =>

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1 o 2 o 3 o .....o 20 = 3

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x o y = ( x-3)(y-3)+3  ?

x o y = xy - 3x - 3y + 12 = x(y-3) -3y + 9 + 3 =

= x(y-3) -3(y-3) + 3 = (x-3)(y-3) + 3 ; (∀) x,y ∈ R