j- Zadanie 9: Kurs maturalny z matematyki rozszerzonej

Rozwiązanie

c) $\frac{x}{( x + 1 ) ( x - 3 )} - \frac{2}{x ^{2} - 9}$

Dziedzina:

$(x + 1) (x - 3) \neq = 0 x + 1 \neq = 0 \land x - 3 \neq = 0 x \neq = - 1 \land x \neq = 3 \land x^{2} - 9 \neq = 0 x^{2} \neq = 9 x \neq = 3 \land x \neq = - 3$

$D = R ∖ {- 3, - 1, 3}$

$\frac{x}{( x + 1 ) ( x - 3 )} - \frac{2}{x ^{2} - 9} = \frac{x ( x + 3 )}{( x + 1 ) ( x - 3 ) ( x + 3 )} - \frac{2 ( x + 1 )}{( x - 3 ) ( x + 3 ) ( x + 1 )} =$
$= \frac{x ^{2} + 3 x}{( x + 1 ) ( x - 3 ) ( x + 3 )} - \frac{2 x + 2}{( x - 3 ) ( x + 3 ) ( x + 1 )} = \frac{x ^{2} + 3 x - 2 x - 2}{( x - 3 ) ( x + 3 ) ( x + 1 )} =$
$= \frac{x ^{2} + x - 2}{( x - 3 ) ( x + 3 ) ( x + 1 )} = \frac{( x + 2 ) ( x - 1 )}{( x ^{2} - 9 ) ( x + 1 )}$

$Δ = 1^{2} - 4 \cdot 1 \cdot (- 2) = 1 + 8 = 9 \to Δ = 9 = 3$
$x_{1} = \frac{- 1 - 3}{2 \cdot 1} = \frac{- 4}{2} = - 2$ oraz $x_{2} = \frac{- 1 + 3}{2 \cdot 1} = \frac{2}{2} = 1$
Więc: $x^{2} + x - 2 = (x + 2) (x - 1)$