Question

Как это решить с помощью метода рационализации?

Answer 1

$\frac{log_3(1-2x-x^2)}{log_{3-\sqrt{5}}(x+1+\sqrt{2})}} \geq 0\\\frac{(3-1)(1-2x-x^2-1)}{(3-\sqrt{5}-1)(x+1+\sqrt{2}-1)} \geq 0\\\frac{-x(x-2)}{x+\sqrt{2}} \leq 0\\\frac{x(x-2)}{x+\sqrt{2}} \geq 0\\-----(-\sqrt{2})++++[0]----[2]++++++>x\\x \in (-\sqrt{2}; 0][2; +\infty)\\\left \{ {{1-2x-x^2 > 0} \atop {x+1+\sqrt{2} > 0 \atop {x+1+\sqrt{2} \neq 1}} \right.\\\left \{ {{x^2 + 2x - 1 < 0} \atop {x> -1 -\sqrt{2} \atop {x\neq-\sqrt{2}}} \right.$

$\left \{ {{(x-(-1+\sqrt{2}))(x-(-1-\sqrt{2})) < 0} \atop {x> -1 - \sqrt{2} \atop {x\neq-\sqrt{2}}} \right.\\\left \{ {{x \in (-1-\sqrt{2};-1+\sqrt{2})} \atop {x> -1 - \sqrt{2}  \atop {x\neq-\sqrt{2}}} \right.\\x \in (-1-\sqrt{2};-\sqrt{2})(-\sqrt{2}; \sqrt{2} - 1)$

Объединяя с решением выше, получим:

$x \in (-\sqrt{2}; 0]$

Ответ: $x \in (-\sqrt{2}; 0]$