Đáp án:
\(2a + b = 6 + \sqrt 7 \)
Giải thích các bước giải:
\(\begin{array}{l}
\sqrt {2{x^2} - 6x + 1} < x - 2\\
\to \left\{ \begin{array}{l}
2{x^2} - 6x + 1 \ge 0\\
x - 2 \ge 0\\
2{x^2} - 6x + 1 < {x^2} - 4x + 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left( { - \infty ;\frac{{3 - \sqrt 7 }}{2}} \right] \cup \left[ {\frac{{3 + \sqrt 7 }}{2}; + \infty } \right)\\
x \ge 2\\
{x^2} - 2x - 3 < 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \in \left( { - \infty ;\frac{{3 - \sqrt 7 }}{2}} \right] \cup \left[ {\frac{{3 + \sqrt 7 }}{2}; + \infty } \right)\\
x \ge 2\\
x \in \left( { - 1;3} \right)
\end{array} \right.\\
KL:x \in \left[ {\frac{{3 + \sqrt 7 }}{2};3} \right)\\
\to a = \frac{{3 + \sqrt 7 }}{2};b = 3\\
\to 2a + b = 6 + \sqrt 7
\end{array}\)
