Odpowiedzi

2009-12-28T00:04:24+01:00
(2x+1)/(x+3) -(x-1)/(x²-9) = (x+3)/(3-x) - (4+x)/(3+x)

D: x²-9≠0
x²≠9
x≠3
x≠-3

(2x+1)/(x+3) -(x-1)/(x²-9) = (x+3)/(3-x) - (4+x)/(3+x) /*(x-3)(x+3)
(2x+1)(x-3)-(x-1)=-(x+3)(x+3)-(x+4)(x-3)
2x²-6x+x-3-x+1=-x²-6x-9-x²+3x-4x+12
2x²+x²+x²-6x+7x-2-3=0
4x²+x-5=0
Δ=1-4*4*(-5)=1+80=81
√Δ=9
x₁=(-1-9)/8=-10/8=-5/4
x₂=(-1+9)/8=8/8=1

Odp. x₁=-5/4 v x₂=1

2009-12-28T00:12:54+01:00
2009-12-28T01:33:07+01:00
[(2x+1)/(x+3)] - [(x-1)/(x²-9)] = [(x+3)/(3-x)] - [(4+x)/(3+x)]
Wyznaczmy najpierw dziedzinę funkcji:
x²-9≠0
x²≠9
x≠3
x≠-3

x∈R - {-3}; {3}

(2x+1)/(x+3) -(x-1)/(x²-9) = (x+3)/(3-x) - (4+x)/(3+x) |*(x-3)(x+3)
(2x+1)(x-3)-(x-1)=-(x+3)(x+3)-(x+4)(x-3)
2x²-6x+x-3-x+1=-x²-6x-9-x²+3x-4x+12
2x²+x²+x²-6x+7x-2-3=0
4x²+x-5=0

Δ=81

x1=(-1-9)/8=-10/8=-5/4
x2=(-1+9)/8=8/8=1