Đáp án:
\(\begin{array}{l}
i)\left\{ \begin{array}{l}
4 \ge x \ge 2\\
x \ne 3
\end{array} \right.\\
k)x > 3\\
l) - \dfrac{1}{2} < x \le 4 - \sqrt 2 \\
m)x \ne 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
i)DK:\left\{ \begin{array}{l}
x - 2 \ge 0\\
4 - x \ge 0\\
\sqrt {4 - x} - 1 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4 \ge x \ge 2\\
4 - x \ne 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4 \ge x \ge 2\\
x \ne 3
\end{array} \right.\\
k)DK:\left\{ \begin{array}{l}
x - 3 \ge 0\\
{x^2} - 9 > 0\\
x \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x > 3\\
\left[ \begin{array}{l}
x > 3\\
x < - 3
\end{array} \right.\\
x \ne 0
\end{array} \right.\\
\to x > 3\\
l)DK:\left\{ \begin{array}{l}
16 - {x^2} \ge 0\\
2x + 1 > 0\\
{x^2} - 8x + 14 \ge 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
16 \ge {x^2}\\
x > - \dfrac{1}{2}\\
{x^2} - 8x + 16 \ge 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4 \ge x \ge - 4\\
x > - \dfrac{1}{2}\\
{\left( {x - 4} \right)^2} \ge 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4 \ge x > - \dfrac{1}{2}\\
\left[ \begin{array}{l}
x - 4 \ge \sqrt 2 \\
x - 4 \le - \sqrt 2
\end{array} \right.
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4 \ge x > - \dfrac{1}{2}\\
\left[ \begin{array}{l}
x \ge 4 + \sqrt 2 \\
x \le 4 - \sqrt 2
\end{array} \right.
\end{array} \right.\\
\to - \dfrac{1}{2} < x \le 4 - \sqrt 2 \\
m)DK:{x^2} - 2x + 1 > 0\\
\to {\left( {x - 1} \right)^2} > 0\\
\to x \ne 1
\end{array}\)
