Rozwiąż równanie :

[tex] \frac{x}{x-1} + \frac{x+1}{x-3} = \frac{ x^{2} -2x-1}{ x^{2}-4x+3 } [/tex]

założenia:
x ≠ 3
x ≠ 1

[tex]
\frac{x}{x - 1} + \frac{x + 1}{x - 3} = \frac{x^{2} - 2x - 1}{x^{2} - 4x + 3}

\frac{x(x - 3) + (x + 1)(x - 1)}{(x - 1)(x - 3)} = \frac{x^{2} - 2x - 1}{(x - 3)(x - 1)}

x(x - 3) + (x + 1)(x - 1) = x^{2} - 2x - 1

x^{2} - 3x + x^{2} - 1 = x^{2} - 2x - 1

x^{2} - 3x = - 2x

x(x - 1) = 0

x = 0 ∨ x = 1

zgodnie z założeniami drugi wynik odrzucam:
x = 0

Answer Link

Анонимviz Анонимviz · Answer 2

\frac{x}{x-1} + \frac{x+1}{x-3} = \frac{ x^{2} -2x-1}{ x^{2}-4x+3 }
x/x-1 + x+1/x-3= (x²-2x-1)/(x²-4x+3)
zal: x rozne od 1, x rozne od 3
x²-4x+3
Δ=16-4*1*3=16-12=4
√Δ=2
x1=4-2/2=2/2=1
x2=4+2/2=6/2=3

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

x(x-3)----------- (x+1)(x-1)------(x²-2x-1)
-------------+ -----------------= -----------------
(x-1)(x-3)-------(x-1)(x-3)-----(x-1)(x-3)

x²-3x------------- x²-x+x-1------(x²-2x-1)
-------------+ -----------------= -----------------
(x-1)(x-3)-------(x-1)(x-3)------(x-1)(x-3)

x²-3x+x²-x+x-1-------(x²-2x-1)
-----------------= -------------------
(x-1)(x-3)-------------(x-1)(x-3)

2x²-3x-1---------------(x²-2x-1)
-----------------= -------------------
(x-1)(x-3)-------------(x-1)(x-3)

2x²-3x-1
Δ=9-4*2*(-1)=9+8=17
√Δ=√17
x1=3-√17/4
x2=3+√17/4

x²-2x-1
Δ=4-4*1*(-1)=4+4=8
√Δ=2√2
x1=2-2√2/2=2(1-√2)/2=1-√2
x2=1+√2