0;\\\frac{1}{t}+t+3-12\leq 0;" alt="\displaystyle(\frac{1}{3})^{-x}+3^x+3 \leq 12;\\\frac{1}{3^x}+3^x+3 \leq 12;\ t=3^x;\ t>0;\\\frac{1}{t}+t+3-12\leq 0;" align="absmiddle" class="latex-formula">
1+t²-9t ≤ 0;
t²-9t+1=0; решим квадратное уравнение
D=81-4*1=77;
t=(9+√77)/2;
t=(9-√77)/2;