Tìm điều kiện của mỗi phương trình sau:
a) \(\sqrt { - 3x + 2} = \frac{2}{{x + 1}}\)
b) \(\sqrt {x - 2} + x = 3{x^2} + 1 - \sqrt { - x - 4} \)
c) \(\frac{{x+4}}{{\sqrt {3{x^2} + 6x + 11} }} = \sqrt {2x + 1} \)
d) \(\frac{{\sqrt { - 3x + 2} }}{{{x^2} - 9}} = x + 2\)
a) ĐKXĐ: \(\left\{ \begin{array}{l}
- 3x + 2 \ge 0\\
x + 1 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le \frac{2}{3}\\
x \ne - 1
\end{array} \right.\)
b) ĐKXĐ: \(\left\{ \begin{array}{l}
x - 2 \ge 0\\
- x - 4 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 2\\
x \le - 4
\end{array} \right. \Leftrightarrow x \in \emptyset \)
c) ĐKXĐ: \(\left\{ \begin{array}{l}
3{x^2} + 6x + 11 > 0\\
2x + 1 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \in R\\
x \ge - \frac{1}{2}
\end{array} \right. \Leftrightarrow x \ge - \frac{1}{2}\)
d) ĐKXĐ: \(\left\{ \begin{array}{l}
x + 4 \ge 0\\
{x^2} - 9 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge - 4\\
x \ne \pm 3
\end{array} \right.\)
-- Mod Toán 10