zad.6.18.

a) \frac{x^2 - 5x + 6}{x^2 - 4x + 3} = 0 \frac{x^2 - 2x - 3x + 6}{x^2 - 3x - x + 3} = 0 \frac{x(x - 2) - 3(x - 2)}{x(x - 3) - (x - 3)} = 0 \frac{(x - 3)(x - 2)}{(x - 1)(x - 3)} = 0 \frac{x - 2}{x - 1} = 0 x - 2 = 0 x = 2 b) \frac{x^3 - 1}{x^2 - 2x + 1} = 0 \frac{(x - 1)(x^2 + x + 1)}{(x - 1)^2} = 0 \frac{x^2 + x + 1}{x - 1} = 0 x^2 + x + 1 = 0 \Delta = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 < 0 Nie ma rozwiązań. c) \frac{x^3 + 5x^2 - 6x}{(x + 1)(x + 4)} = 0 \frac{x(x^2 + 5x - 6)}{(x + 1)(x + 4)} = 0 \frac{x(x^2 + 6x - x - 6)}{(x + 1)(x + 4)} = 0 \frac{x(x(x + 6) - (x + 6))}{(x + 1)(x + 4)} = 0 \frac{x(x - 1)(x + 6)}{(x + 1)(x + 4)} = 0 x = 0 lub x = 1 lub x = -6