Question

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a) x^2+5x+6 dzielone na x^2-4

b) x^2-8x+15 dzielone na 2x^2-50

Answer 1

a)

x²+5x+6 / x²- 4

 x² - 4 ≠ 0

(x - 2)(x + 2) ≠ 0

x ≠ 2 i x ≠ - 2

x ∈ R\{±2} 

 x²+5x+6 = 0

Δ = 25 - 24 = 1

√Δ = 1

x₁ = (- 5 - 1)/2 = - 6/2 = - 3

x₂ = (- 5 + 1)/2 = - 4/2 = - 2

 

x²+5x+6 / x²- 4

(x + 3)(x + 2)/(x - 2)(x + 2)

x + 3/x + 2 

 

b)

x²-8x+15 / 2x²-50 

2x² - 50 ≠ 0

x² - 25 ≠ 0

x ≠ ±√25

x ≠ ± 5

x ∈ R \ {± 5}

 x²-8x+15 = 0

Δ = 64 - 60 = 4

√Δ = √4 = 2

x₁ = (8 - 2)/2 = 6/2 = 3

x₂ = (8 + 2)/2 = 10/2 = 5

 

(x - 3)(x - 5)/2(x - 5)(x + 5) = (x - 3)/2(x + 5) 

 

 

Answer 2

a)   x²+5x+6=0

Δ=b²-4ac

Δ=25-24

Δ=1

√Δ=1

x 1=(-5-1)/2            x 1=-3

x 2=(-5+1)/2           x 2=-2

x²+5x+6=(x+2)(x+3)

[(x+2)(x+3)]/(x+2)(x-2)=(x+3)/(x-2)

zalozenia:      x²-4≠o

                       (x+2)(x-2)≠0

                       x≠-2    i    x≠2

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b)   x²-8x+15=0

Δ=64-60

Δ=4

√Δ=2

x 1=(8-2)/2       x 1=3

x 2= (8+2)/2         x 2=5

x²-8x+15=(x-3)(x-5)

2x²-50=2(x²-25)=2(x+5)(x-5)

[(x-3)(x-5)]/[2(x+5)(x-5)]=(x-3)/{2(x+5)]

zalozenia   2x²-50≠0

                x²-25≠0

                (x+5)(x-5)≠0

                x≠-5  i x≠5