Question

Решите пожалуйста неравенство !!!!

Answer 1

$3^x+\frac{8}{3^x} \geq 9 ; 3^x \ \textgreater \ 0\\ 3^{2x} +8 \geq 9 * 3^x \\ 3^{2x}-9*x^x+8 \geq 0\\D =81-32=49\\ 3^x=\frac{9 \pm 7}{2} = 1;8 \\ \\ \left \{ {{3^x \leq 1} \atop {3^x \geq 8}} \right. \\ \left \{ {{x \leq 0} \atop {x \geq\log_{3}8}} \right. \\\\ x \in (-\infty;0] \cup[\log_{3}8; +\infty)$

Answer 2

3^x+8/3^x-9≥0
3^2x-9*3^x+8≥0
3^x=a
a²-9a+8≥0
a1+a2=9 U a1*a2=8⇒a1=1 U a2=8
a≤1⇒3^x≤1⇒x≤0
a≥8⇒3^x≥8⇒x≥log(3)8
x∈(-∞;0] U [log(3)8;∞)