Đáp án:
\[\left[ \begin{array}{l}
x > 7 + 4\sqrt 3 \\
0 \le x < 3
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ: \(x \ge 0\)
Ta có:
\(\begin{array}{l}
x + \sqrt x > \left( {2\sqrt 3 + 3} \right)\left( {\sqrt x - 1} \right)\\
\Leftrightarrow x + \sqrt x > \left( {2\sqrt 3 + 3} \right)\sqrt x - \left( {2\sqrt 3 + 3} \right)\\
\Leftrightarrow {\left( {\sqrt x } \right)^2} - \left( {2\sqrt 3 + 2} \right)\sqrt x + \left( {2\sqrt 3 + 3} \right) > 0\\
\Leftrightarrow \left( {\sqrt x - \sqrt 3 } \right)\left( {\sqrt x - \left( {2 + \sqrt 3 } \right)} \right) > 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x > 2 + \sqrt 3 \\
\sqrt x < \sqrt 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x > 7 + 4\sqrt 3 \\
0 \le x < 3
\end{array} \right.
\end{array}\)
