Đáp án: $x = \dfrac{{10}}{3};x = \dfrac{8}{3}$
Giải thích các bước giải:
$\begin{array}{l}
f\left( x \right) = 3{x^2} - 18x + 27\\
g\left( x \right) = \left| {x - 3} \right|\\
h\left( x \right) = f\left( x \right) - 2g\left( x \right) + 1\\
= 3{x^2} - 18x + 27 - 2\left| {x - 3} \right| + 1\\
= 3\left( {{x^2} - 6x + 9} \right) - 2\left| {x - 3} \right| + 1\\
= 3.{\left( {x - 3} \right)^2} - 2\left| {x - 3} \right| + 1\\
= 3.{\left( {\left| {x - 3} \right|} \right)^2} - 2\left| {x - 3} \right| + 1\\
3.\left[ {{{\left( {\left| {x - 3} \right|} \right)}^2} - \dfrac{2}{3}\left| {x - 3} \right|} \right] + 1\\
= 3.\left[ {{{\left( {\left| {x - 3} \right|} \right)}^2} - 2.\left| {x - 3} \right|.\dfrac{1}{3} + \dfrac{1}{9}} \right] - 3.\dfrac{1}{9} + 1\\
= 3.{\left( {\left| {x - 3} \right| - \dfrac{1}{3}} \right)^2} + \dfrac{2}{3} \ge \dfrac{2}{3}\\
\Leftrightarrow h\left( x \right) \ge \dfrac{2}{3}\\
\Leftrightarrow GTNN:h\left( x \right) = \dfrac{2}{3}\,\\
Khi:\left| {x - 3} \right| = \dfrac{1}{3} \Leftrightarrow \left[ \begin{array}{l}
x - 3 = \dfrac{1}{3} \Leftrightarrow x = \dfrac{{10}}{3}\\
x - 3 = \dfrac{{ - 1}}{3} \Leftrightarrow x = \dfrac{8}{3}
\end{array} \right.\\
Vay\,x = \dfrac{{10}}{3};x = \dfrac{8}{3}
\end{array}$
