Giải thích các bước giải:
Ta có :
$A=3^{2x}+3^{2x-1}+..+3^{2x-9}$
$\to 3A=3^{2x+1}+3^{2x}+..+3^{2x-8}$
$\to 3A-A=3^{2x+1}-3^{2x-9}$
$\to 2A=3^{2x+1}-3^{2x-9}$
$\to A=\dfrac{1}{2}(3^{2x+1}-3^{2x-9})$
$\to \dfrac{1}{2}(3^{2x+1}-3^{2x-9})<61.3^{-5}$
$\to 3^{2x+1}-3^{2x-9}<122.3^{-5}$
$\to 3^{2x}.3-3^{2x}.3^{-9}<122.3^{-5}$
$\to 3^{2x}(3-3^{-9})<122.3^{-5}$
$\to 3^{2x}<\dfrac{122.3^{-5}}{3-3^{-9}}$
$\to 2x<\log_3(\dfrac{122.3^{-5}}{3-3^{-9}})$
