Đáp án:
d. \(\left[ \begin{array}{l}
x \ge \sqrt 3 \\
x \le - \sqrt 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
b.DK:\left\{ \begin{array}{l}
\dfrac{{x - 4}}{{3 - 2x}} \ge 0\\
3 - 2x \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 4 \ge 0\\
3 - 2x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 4 \le 0\\
3 - 2x < 0
\end{array} \right.
\end{array} \right.\\
x \ne \dfrac{3}{2}
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 4\\
\dfrac{3}{2} > x
\end{array} \right.\left( l \right)\\
\left\{ \begin{array}{l}
x \le 4\\
\dfrac{3}{2} < x
\end{array} \right.
\end{array} \right.\\
\to \dfrac{3}{2} < x \le 4\\
c.DK:4{x^2} + 3x \ge 0\\
\to x\left( {4x + 3} \right) \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \ge 0\\
4x + 3 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x \le 0\\
4x + 3 \le 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x \ge 0\\
x \le - \dfrac{3}{4}
\end{array} \right.\\
d.DK:{x^2} - 3 \ge 0\\
\to {x^2} \ge 3\\
\to \left[ \begin{array}{l}
x \ge \sqrt 3 \\
x \le - \sqrt 3
\end{array} \right.\\
e.DK:{x^2} - 4x \ne 0\\
\to x \ne \left\{ {0;4} \right\}
\end{array}\)
