Đáp án:
\(x \in \left( { - \infty ;\dfrac{1}{3}} \right) \cup \left( {3; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:{x^2} + 4x + 4 \ne 0\\
\to {\left( {x + 2} \right)^2} \ne 0\\
\to x + 2 \ne 0\\
\to x \ne - 2\\
\dfrac{{3{x^2} - 10x + 3}}{{{{\left( {x + 2} \right)}^2}}} \ge 0\\
\to \dfrac{{\left( {x - 3} \right)\left( {3x - 1} \right)}}{{{{\left( {x + 2} \right)}^2}}} \ge 0\\
\to \left( {x - 3} \right)\left( {3x - 1} \right) \ge 0\left( {do:{{\left( {x + 2} \right)}^2} > 0\forall x \ne - 2} \right)\\
\to x \in \left( { - \infty ;\dfrac{1}{3}} \right) \cup \left( {3; + \infty } \right)
\end{array}\)
