Đáp án:
8B) b) \( - \dfrac{2}{3} \le x \le \dfrac{2}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
8A)\\
a)DK:{x^2} - 8x - 9 \ge 0\\
\to {x^2} - 8x + 16 - 7 \ge 0\\
\to {\left( {x - 4} \right)^2} \ge 7\\
\to \left[ \begin{array}{l}
x - 4 \ge \sqrt 7 \\
x - 4 \le - \sqrt 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
x \ge 4 + \sqrt 7 \\
x \le 4 - \sqrt 7
\end{array} \right.\\
b)DK:\dfrac{{2x - 4}}{{5 - x}} \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x - 4 \ge 0\\
5 - x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - 4 \le 0\\
5 - x < 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 2\\
5 > x
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 2\\
x > 5
\end{array} \right.\left( l \right)
\end{array} \right.\\
8B)a)DK:\dfrac{{x - 6}}{{x - 2}} \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 6 \ge 0\\
x - 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 6 \le 0\\
x - 2 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 6\\
x > 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 6\\
x < 2
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x \ge 6\\
x < 2
\end{array} \right.\\
b)DK:4 - 9{x^2} \ge 0\\
\to 9{x^2} \le 4\\
\to - 2 \le 3x \le 2\\
\to - \dfrac{2}{3} \le x \le \dfrac{2}{3}
\end{array}\)
