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2009-10-10T14:25:47+02:00
A)
(x + 3)/(x² + 6x + 9) - x/(x² + 3x)

założenia:
x² + 6x + 9 ≠ 0
(x + 3)² ≠ 0
x ≠ -3

x² + 3x ≠ 0
x(x + 3) ≠ 0
x ≠ -3 i x ≠0

(x + 3)/(x² + 6x + 9) - x/(x² + 3x) = (x + 3)/[(x + 3)²] - x/[x(x + 3)] = 1/(x + 3) - 1/(x + 3) = 0

b)
(x + 7)/(x² + x - 6) + x/(x² - 9) - 2/(2x - 4)

założenia:
x² + x - 6 ≠ 0
(x - 2)(x + 3) ≠ 0
x ≠ 2 i x ≠ -3

x² - 9 ≠ 0
(x - 3)(x + 3) ≠ 0
x ≠ 3 i x ≠ -3

2x - 4 ≠ 0
x ≠ 2

(x + 7)/(x² + x - 6) + x/(x² - 9) - 2/(2x - 4) = (x + 7)/[(x - 2)(x + 3)] + x/[(x - 3)(x + 3)] - 2/2(x - 2) = (x + 7)/[(x - 2)(x + 3)] + x/[(x - 3)(x + 3)] - 1/(x - 2) = [(x + 7)(x - 3)]/[(x - 2)(x + 3)(x - 3)] + [x(x - 2)]/[(x - 3)(x + 3)(x - 2)] - [(x + 3)(x - 3)]/[(x - 2)(x + 3)(x - 3)] = [(x + 7)(x - 3) + x(x - 2) - (x + 3)(x - 3)]/[(x - 2)(x + 3)(x - 3)] = [x² + 7x - 3x - 21 + x² - 2x - x² + 9]/[(x - 2)(x + 3)(x - 3)] = [2x - 12 + x²]/[(x - 2)(x + 3)(x - 3)] = [x² + 2x - 12]/[(x - 2)(x + 3)(x - 3)]
1 5 1
2009-10-10T14:46:29+02:00
(x + 3)/(x² + 6x + 9) - x/(x² + 3x)=

zał.(x² + 6x + 9)≠0 i(x² + 3x)≠0
(x+3)²≠0 i x(x+3)≠0
x≠-3 i x≠0 i x≠-3
D=R\{-3,0}
(x + 3)/(x+3)²- x/x(x + 3)=
=1/(x+3) -1/(x+3)=0

(x + 7)/(x² + x - 6) + x/(x² - 9) - 2/(2x - 4)=

zał.(x² + x - 6) ≠0 i (x² - 9)≠0 i 2x - 4)≠0

x² + x - 6 ≠ 0 (x-3)(x+3)≠0 x≠2
delta=25

x ≠ 2 i x ≠ -3 x≠3 i x≠-3
D=R\{-3,2,3}

(x + 7)/(x² + x - 6) + x/(x² - 9) - 2/(2x - 4) =
=(x + 7)/[(x - 2)(x + 3)] + x/[(x - 3)(x + 3)] - 2/2(x - 2) =
=(x + 7)/[(x - 2)(x + 3)] + x/[(x - 3)(x + 3)] - 1/(x - 2) =
= [(x + 7)(x - 3)]/[(x - 2)(x + 3)(x - 3)]+[x(x - 2)]/[(x - 3)(x + 3)(x - 2)]-[(x + 3)(x - 3)]/[(x - 2)(x + 3)(x - 3)] =
=[x² + 7x - 3x - 21 + x² - 2x - x² + 9]/[(x - 2)(x + 3)(x - 3)] =
= (2x - 12 + x²)/[(x - 2)(x + 3)(x - 3)] =
=([x² + 2x - 12)/[(x - 2)(x + 3)(x - 3)]