Đáp án:
10) \(x \ge \dfrac{1}{6)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:3x - 1 \ge 0\\
\to x \ge \dfrac{1}{3}\\
2)Do:{x^2} + 3 > 0\forall x \in R\\
\to DK:\forall x\\
3)DK:5 - 2x \ge 0\\
\to x \le \dfrac{5}{2}\\
4)DK:{x^2} - 2 \ge 0\\
\to \left[ \begin{array}{l}
x \ge \sqrt 2 \\
x \le - \sqrt 2
\end{array} \right.\\
5)DK:7x - 14 > 0\\
\to x > 2\\
6)DK:{x^2} - 3x + 7 \ge 0\\
\to {x^2} - 2x.\dfrac{3}{2} + \dfrac{9}{4} + \dfrac{{19}}{4} \ge 0\\
\to {\left( {x - \dfrac{3}{2}} \right)^2} + \dfrac{{19}}{4} \ge 0\left( {ld} \right)\forall x \in R\\
\to DK:\forall x\\
7)DK:2x - 1 \ge 0 \to x \ge \dfrac{1}{2}\\
8)DK:{x^2} - 9 \ge 0\\
\to \left( {x - 3} \right)\left( {x + 3} \right) \ge 0\\
\to \left[ \begin{array}{l}
x \ge 3\\
x \le - 3
\end{array} \right.\\
9)DK:\left\{ \begin{array}{l}
\dfrac{{x + 3}}{{7 - x}} \ge 0\\
x \ne 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 3 \ge 0\\
7 - x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 3 \le 0\\
7 - x < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge - 3\\
x < 7
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le - 3\\
x > 7
\end{array} \right.\left( l \right)
\end{array} \right.\\
\to - 3 \le x < 7\\
10)DK:\left\{ \begin{array}{l}
6x - 1 \ge 0\\
x + 3 \ge 0
\end{array} \right. \to \left\{ \begin{array}{l}
x \ge \dfrac{1}{6}\\
x \ge - 3
\end{array} \right.\\
\to x \ge \dfrac{1}{6}
\end{array}\)
