Đáp án:
a. \(\left[ \begin{array}{l}
x \ge \dfrac{1}{2}\\
x \le - \dfrac{1}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a.DK:4{x^2} - 1 \ge 0\\
\to \left( {2x - 1} \right)\left( {2x + 1} \right) \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
2x - 1 \ge 0\\
2x + 1 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
2x - 1 \le 0\\
2x + 1 \le 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x \ge \dfrac{1}{2}\\
x \le - \dfrac{1}{2}
\end{array} \right.\\
b.DK:2{x^2} + 4x + 5 \ge 0\\
\to 2{x^2} + 2.x\sqrt 2 .\sqrt 2 + 2 + 3 \ge 0\\
\to {\left( {x\sqrt 2 + \sqrt 2 } \right)^2} + 3 \ge 0\left( {ld} \right)\forall x \in R\\
c.DK: - {x^2} + 2x > 0\\
\to - x\left( {x - 2} \right) > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
- x > 0\\
x - 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
- x < 0\\
x - 2 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x < 0\\
x > 2
\end{array} \right.\left( l \right)\\
\left\{ \begin{array}{l}
x > 0\\
x < 2
\end{array} \right.
\end{array} \right.\\
d.DK:\left\{ \begin{array}{l}
x \ne 0\\
- 3x > 0\\
x + \dfrac{3}{x} \ge 0
\end{array} \right. \to \left\{ \begin{array}{l}
x \ne 0\\
x < 0\\
\dfrac{{{x^2} + 3}}{x} \ge 0
\end{array} \right.\left( {vô lý} \right)
\end{array}\)
⇒ Không tồn tại x để biểu thức xác định
