Đáp án:
c. \(x \in \left( {\dfrac{5}{3}; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
a.2{x^2} + 3x \ge 3\sqrt {2{x^2} + 3x + 9} + 9\\
\to 2{x^2} + 3x - 9 \ge 3\sqrt {2{x^2} + 3x + 9} \left( 1 \right)\\
Đặt:\sqrt {2{x^2} + 3x + 9} = t\left( {t \ge 0} \right)\\
\to 2{x^2} + 3x + 9 = {t^2}\\
\to 2{x^2} + 3x = {t^2} - 9\\
\left( 1 \right) \to {t^2} - 9 - 9 = 3t\\
\to {t^2} - 3t - 18 = 0\\
\to \left( {t - 6} \right)\left( {t + 3} \right) = 0\\
\to \left[ \begin{array}{l}
t = 6\\
t = - 3\left( l \right)
\end{array} \right.\\
\to \sqrt {2{x^2} + 3x + 9} = 6\\
\to 2{x^2} + 3x + 9 = 36\\
\to \left[ \begin{array}{l}
x = 3\\
x = - \dfrac{9}{2}
\end{array} \right.\\
b.\sqrt { - {x^2} + 4x - 3} < 2x - 5\\
\to \left\{ \begin{array}{l}
- {x^2} + 4x - 3 \ge 0\\
2x - 5 \ge 0\\
- {x^2} + 4x - 3 < 4{x^2} - 20x + 25
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left[ {1;3} \right]\\
x \ge \dfrac{5}{2}\\
5{x^2} - 24x + 28 > 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left[ {\dfrac{5}{2};3} \right]\\
x \in \left( { - \infty ;2} \right) \cup \left( {\dfrac{{14}}{5}; + \infty } \right)
\end{array} \right.\\
\to x \in \left( {\dfrac{{14}}{5};3} \right]\\
c.\left| {x - 2} \right| < 2x - 3\\
\to \left\{ \begin{array}{l}
{x^2} - 4x + 4 < 4{x^2} - 12x + 9\\
2x - 3 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
3{x^2} - 8x + 5 > 0\\
x \ge \dfrac{3}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left( { - \infty ;1} \right) \cup \left( {\dfrac{5}{3}; + \infty } \right)\\
x \ge \dfrac{3}{2}
\end{array} \right.\\
\to x \in \left( {\dfrac{5}{3}; + \infty } \right)
\end{array}\)
