Question

СРОЧНО!! Решите систему неравенств. 11 КЛАСС

Answer 1

Ответ: x∈(0,5;2).

Объяснение:

$\displaystyle\\\left \{ {{2^{x^2}*0,25 < (\frac{1}{4})^{-1}} \atop {\frac{9^{x+3}}{27}} > (\frac{1}{9})^{-2} }} \right. \ \ \ \ \ \ \left \{ {{2^{x^2}*\frac{1}{4} < 4^1} \atop {\frac{3^{2*(x+3)}}{3^3} > 9^2}}} \right.\ \ \ \ \ \ \left \{ {{\frac{2^{x^2}}{2^2} } < 2^2} \atop {\frac{3^{2x+6}}{3^3} > 81} \right. \ \ \ \ \ \ \left \{ {{2^{x^2-2} < 2^2} \atop {3^{2x+3} > 3^4}} \right.$

$\displaystyle\\\left \{ {{x^2-2 < 2} \atop {2x+3 > 4}} \right. \ \ \ \ \ \ \left \{ {{x^2-4 < 0} \atop {2x > 1\ |:2}} \right. \ \ \ \ \ \ \left \{ {{(x+2)*(x-2) < 0} \atop {x > 0,5}} \right. \ \ \ \ \ \ \left \{ {{x\in(-2;2)} \atop {x\in(0,5;+\infty)}} \right. .$

$x\in(0,5;2).$