Question

ajutor îmi trebu urgent ​

Answer 1

Explicație pas cu pas:

4.a)

x² - 4 ≠ 0 => x ≠ ±2

$\dfrac{1}{ {x}^{2} - 4} + \dfrac{^{x + 2)} 1}{x - 2} - \dfrac{^{x - 2)}1}{x + 2} = \dfrac{1 + (x + 2) - (x - 2)}{(x - 2)(x + 2)} = \dfrac{1 + x + 2 - x + 2}{(x - 2)(x + 2)} = \dfrac{5}{(x - 2)(x + 2)}$

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

b)

x ≠ 0

x³+2x = x(x²+2) ≠ 0 => x ≠0

2-x ≠ 0 => x ≠ 2

$\dfrac{4 - {x}^{2} }{x} : \dfrac{2 - x}{ {x}^{3} + 2x} = \dfrac{(2 - x)(2 + x)}{x} \cdot \dfrac{x( {x}^{2} + 2)}{2 - x} = ( {x}^{2} + 2)(x + 2)$

$x \in \mathbb{R}{/}\{0;2\}$