Đáp án:
\(0 \le x < 4\)
Giải thích các bước giải:
\(\begin{array}{l}
P = \dfrac{{x + 2\sqrt x + x - 2\sqrt x - x + 2\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
M = P:Q = \dfrac{{\sqrt x }}{{\sqrt x - 2}}:\dfrac{{\sqrt x + 2}}{{\sqrt x - 2}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 2}}.\dfrac{{\sqrt x - 2}}{{\sqrt x + 2}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 2}}\\
{M^2} < \dfrac{1}{4}\\
\to - \dfrac{1}{2} < M < \dfrac{1}{2}\\
\to \left\{ \begin{array}{l}
- \dfrac{1}{2} < \dfrac{{\sqrt x }}{{\sqrt x + 2}}\left( {ld} \right)\left( {do:\left\{ \begin{array}{l}
\sqrt x \ge 0\forall x \ge 0\\
\sqrt x + 2 > 0
\end{array} \right.} \right)\\
\dfrac{{\sqrt x }}{{\sqrt x + 2}} < \dfrac{1}{2}
\end{array} \right.\\
\to \dfrac{{2\sqrt x - \sqrt x - 2}}{{2\left( {\sqrt x + 2} \right)}} < 0\\
\to \dfrac{{\sqrt x - 2}}{{2\left( {\sqrt x + 2} \right)}} < 0\\
\to \sqrt x - 2 < 0\left( {do:\sqrt x + 2 > 0\forall x \ge 2} \right)\\
\to x < 4\\
\to 0 \le x < 4
\end{array}\)
