Question

Urgenttt!!
Dau coroana!!​

Answer 1

a)

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

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

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

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

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

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

b)

$\dfrac{1}{E(x)}\leq 0 \iff \dfrac{1}{\dfrac{x + 1}{(x - 1)^{2}}}\leq 0$

$\dfrac{(x - 1)^{2}}{x + 1} \leq 0$

$(x - 1)^{2} \geq 0 \iff x + 1 < 0 \iff x < -1$

$\implies x \in \Big(-\infty;-1\Big)$