Question

F.Wymierna

skróć ułamki podaj konieczne założenia:

a)x³-1/2x²-2
b)3x²+6x+3/x³+1
c)2x³-16/x²+2x+4
d)x²+5x+25/x³-125
e)x²-x-6/x²-4
f)36+3x-3x²/x²+6x+9
g)-2x²-14x+16/3x²+15x-72
h)2x²-3x-9/4x²-11x-3
i)6x²+13x-5/9x²+3x-2

Answer 1

a)

(x³-1)/(2x²-2)

 

Zał. 2x²-2 ≠ 0

2(x²-1) ≠ 0 |:2

x²-1 ≠ 0

(x-1)(x+1)≠0

x-1≠0  ∧  x+1≠0

x≠1  ∧  x≠-1

 

(x³-1)/(2x²-2) = (x-1)(x²+x+1) / 2(x²-1) = (x-1)(x²+x+1) / 2(x-1)(x+1) = (x²+x+1) / 2(x+1)

b)

(3x²+6x+3)/(x³+1)

Zał. x³+1≠ 0

(x+1)(x²-x+1) ≠ 0

x+1 ≠ 0  ∧ x²-x+1 ≠ 0

x ≠ - 1

(3x²+6x+3)/(x³+1) = 3(x²+2x+1) / (x+1)(x²-x+1) = 3(x+1)² / (x+1)(x²-x+1) = 3(x+1) / (x²-x+1)

c)

(2x³-16)/(x²+2x+4)

Zał. x²+2x+4≠0

Δ=4-16=-8<0 (nie ma miejsc zerowych)

x∈R

(2x³-16)/(x²+2x+4) = 2(x³-8)/(x²+2x+4) = 2(x³-2³)/(x²+2x+4) = 2(x-2)(x²+2x+4)/(x²+2x+4) = 2(x-2)

d)

(x²+5x+25)/(x³-125)

Zał. x³-125≠0

x³-5³ ≠ 0

(x-5)(x²+5x+25) ≠ 0

x-5 ≠ 0  ∧ x²+5x+25 ≠ 0

x ≠ 5

(x²+5x+25)/(x³-125) = (x²+5x+25)/(x-5)(x²+5x+25) = 1/(x- 5)

e)

(x²-x-6)/(x²-4)

Zał. x²-4≠0

(x-2)(x+2) ≠ 0

x-2 ≠ 0  ∧ x+2 ≠ 0

x ≠ 2  ∧  x ≠ - 2

 

(x²-x-6)/(x²-4) = (x+2)(x-3) /(x-2)(x+2) = (x-3)/(x-2)

 

_________

x²-x-6

Δ = 1 + 24 = 25; √Δ = 5

x₁ = (1-5) : (2·1) = -4:2 = -2

x₂ = (1+5) : (2·1) = 6:2 = 3

x²-x-6 = (x+2)(x-3)

 

f)

(36+3x-3x²)/(x²+6x+9)

Zał. x²+6x+9 ≠ 0

(x+3)² ≠ 0

x+3 ≠ 0

x ≠ -3

(36+3x-3x²)/(x²+6x+9) = -3(x²-x-12)/(x+3)² = -3(x+3)(x-4)/(x+3)² = -3(x-4)/(x+3)

______

x²-x-12

Δ = 1+48=49; √Δ = 7

x₁ = (1-7):(2·1) = -6 : 2 = -3

x₂ = (1+7):(2·1) = 8 : 2 = 4

x²-x-12 = (x+3)(x-4)

g)

(-2x²-14x+16)/(3x²+15x-72)

Zał. 3x²+15x-72≠0

3(x²+5x-24) ≠ 0

x²+5x-24 ≠ 0

Δ = 25+96=121; √Δ = 11

x₁ = (-5-11):(2·1) = -16 : 2 = -8

x₂ = (-5+11):(2·1) = 6 : 2 = 3

x ≠ - 8  ∧  x ≠ 3

3x²+15x-72 = 3(x+8)(x-3)

 

(-2x²-14x+16)/(3x²+15x-72) = -2(x+8)(x-1) / 3(x+8)(x-3) = -2(x-1) / 3(x-3)

_______

-2x²-14x+16 = -2(x²+7x-8)

x²+7x-8

Δ = 49+32 = 81; √Δ = 9

x₁ = (-7-9):(2·1) = -16 : 2 = -8

x₂ = (-7+9):(2·1) = 2 : 2 = 1

-2x²-14x+16 = -2(x+8)(x-1)

h)

(2x²-3x-9)/(4x²-11x-3)

Zał. 4x²-11x-3≠0

Δ = 121+48=169; √Δ = 13

x₁ = (11-13):(2·4) = -2 : 8 = -¼

x₂ = (11+13):(2·4) = 24 : 8 = 3

x ≠ -¼  ∧ x ≠ 3

4x²-11x-3 = 4(x+¼)(x-3) = (4x+1)(x-3)

 

(2x²-3x-9)/(4x²-11x-3) = (2x+3)(x-3) /(4x+1)(x-3) = (2x+3)/(4x+1)

________

2x²-3x-9

Δ = 9+72=81; √Δ =9

x₁ = (3-9):(2·2) = -6 : 4 = - 1,5

x₂ = (3+9):(2·2) = 12 : 4 = 3

2x²-3x-9 = 2(x+1,5)(x-3) = (2x+3)(x-3)

i)

(6x²+13x-5)/(9x²+3x-2)

Zał. 9x²+3x-2 ≠ 0

Δ = 9+72=81; √Δ = 9

x₁ = (-3-9):(2·9) = -12 : 18 = - ⅔

x₂ = (-3+9):(2·9) = 6 : 18 = ⅓

9x²+3x-2 = 9(x+⅔)(x-⅓) = (3x+2)(3x-1)

 

(6x²+13x-5)/(9x²+3x-2) = (2x+5)(3x-1) / (3x+2)(3x-1) = (2x+5)/(3x+2)

 

_____

6x²+13x-5

Δ = 169+120=289; √Δ = 17

x₁ = (-13-17):(2·6) = -30 : 12 = - 2,5

x₂ = (-13+17):(2·6) = 4 : 12 = ⅓

6x²+13x-5 = 6(x+2,5)(x-⅓) = (2x+5)(3x-1)