Question

(1/3)^x + 3^x+3 ≤ 12
(34 баллов)

Answer 1

$displaystyle(frac{1}{3})^{-x}+3^x+3 leq 12;\frac{1}{3^x}+3^x+3 leq 12; t=3^x; t>0;\frac{1}{t}+t+3-12leq 0;$

1+t²-9t ≤ 0;

t²-9t+1=0; решим квадратное уравнение

D=81-4*1=77;

t=(9+√77)/2;

t=(9-√77)/2;

$displaystyle t=frac{9pmsqrt{77}}{2};\3^x=frac{9pmsqrt{77}}{2};\x=log_3frac{9pmsqrt{77}}{2};$

$displaystyle  + - + \. log_3frac{9-sqrt{77}}{2} log_3frac{9+sqrt{77}}{2}\\x in [log_3frac{9-sqrt{77}}{2};log_3frac{9+sqrt{77}}{2}];$