Question

Simplificati rapoartele si stabiliti domeniul lor de definitie

Answer 1

Explicație pas cu pas:

9.a)

$\dfrac{ {x}^{2} - 9}{ {x}^{2} + 5x + 6} = \dfrac{(x - 3)(x + 3)}{(x + 2)(x + 3)} = \dfrac{x - 3}{x + 2}$

$x \in \mathbb{R} - \{-2;-3\}$

d)

$\dfrac{ {x}^{2} - 5x + 6}{ {x}^{2} - 4x + 4} = \dfrac{(x - 2)(x - 3)}{ {(x - 2)}^{2} } = \dfrac{x - 3}{x - 2}$

$x \in \mathbb{R} - \{2\}$

10.a)

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

$x \in \mathbb{R} - \{-3;1\}$

d)

$\dfrac{ {x}^{2} + x - 6}{ {x}^{2} + 2x - 8} = \dfrac{(x - 2)(x + 3)}{(x - 2)(x + 4)} = \dfrac{x + 3}{x + 4}$

$x \in \mathbb{R} - \{-4;2\}$

11.a)

$\dfrac{ {x}^{2} + 8x + 12}{ {x}^{2} + 5x - 6} = \dfrac{(x + 2)(x + 6)}{(x - 1)(x + 6)} = \dfrac{x + 2}{x - 1}$

$x \in \mathbb{R} - \{-6;1\}$

d)

$\dfrac{ {x}^{2} - 5x + 6}{ {x}^{2} - 7x + 12} = \dfrac{(x - 2)(x - 3)}{(x - 3)(x - 4)} = \dfrac{x - 2}{x - 4}$

$x \in \mathbb{R} - \{3;4\}$

Answer 2

Răspuns:

Explicație pas cu pas: