Question

13-5•(3^х)
————————— ≥0,5
(9^х)-12•(3^х)+27

Answer 1

Ответ:

13–5·3x / (9x–12·3x+27) ≥ 0.5⪻⪼t^2-2t+1/t^2-12t+27≤0⪻⪼(t-1)^2/(t-3)(t-9)≤0 t=1 1<x<2

Answer 2

$\frac{13-5\cdot 3^x}{9^x-12\cdot 3^x+27}\geq 0,5\Leftrightarrow \frac{-5\left ( 3^x-3 \right )}{3^x\left ( 3^x-3 \right )-9\left ( 3^x-3 \right )}\geq 0,5\Leftrightarrow -\frac{5}{3^x-9}\geq 0,5\\\frac{1+3^x}{3^x-9}\leq 0\Rightarrow 3^x\in \left [ -1;9 \right )\Rightarrow x\in \left ( -\infty ;2 \right )\setminus \left \{ 1 \right \}$