Đáp án:
A
Giải thích các bước giải:
$$\eqalign{
& y = \sqrt {{{\log }_{{1 \over 3}}}\left( {x - 3} \right) + 2} \cr
& DKXD:\,\,\left\{ \matrix{
{\log _{{1 \over 3}}}\left( {x - 3} \right) + 2 \ge 0 \hfill \cr
x - 3 > 0 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
{\log _{{1 \over 3}}}\left( {x - 3} \right) \ge - 2 \hfill \cr
x > 3 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
x - 3 \le {\left( {{1 \over 3}} \right)^{ - 2}} \hfill \cr
x > 3 \hfill \cr} \right. \cr
& \Leftrightarrow \left\{ \matrix{
x - 3 \le 9 \hfill \cr
x > 3 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
x \le 12 \hfill \cr
x > 3 \hfill \cr} \right. \cr
& \Rightarrow D = \left( {3;12} \right] \cr
& Chon\,\,A. \cr} $$
