Đáp án:
\(\left[ \begin{array}{l}
0 < x \le 1\\
2 \le x < 10
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
C31:\\
\sqrt {2{x^2} - 6x + 4} < x + 2\\
\to \left\{ \begin{array}{l}
2{x^2} - 6x + 4 \ge 0\\
x + 2 \ge 0\\
2{x^2} - 6x + 4 < {x^2} + 4x + 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2\left( {x - 2} \right)\left( {x - 1} \right) \ge 0\\
x \ge - 2\\
{x^2} - 10x < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
x \le 1
\end{array} \right.\\
x \ge - 2\\
x\left( {x - 10} \right) < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left[ \begin{array}{l}
x \ge 2\\
- 2 \le x \le 1
\end{array} \right.\\
0 < x < 10
\end{array} \right.\\
KL:\left[ \begin{array}{l}
0 < x \le 1\\
2 \le x < 10
\end{array} \right.
\end{array}\)
