Đáp án:
\(0 \le x < 1\)
Giải thích các bước giải:
\(\begin{array}{l}
P = \frac{{\sqrt x + 1 + \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} = \frac{{2\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
Q = \frac{{\sqrt x - \sqrt x + 1}}{{\sqrt x - 1}} = \frac{1}{{\sqrt x - 1}}\\
M = P:Q = \frac{{2\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}:\frac{1}{{\sqrt x - 1}}\\
= \frac{{2\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\left( {\sqrt x - 1} \right) = \frac{{2\sqrt x + 1}}{{\sqrt x + 1}}\\
Do:M < \frac{3}{2} \to \frac{{2\sqrt x + 1}}{{\sqrt x + 1}} < \frac{3}{2}\\
\to \frac{{4\sqrt x + 2 - 3\sqrt x - 3}}{{2\left( {\sqrt x + 1} \right)}} < 0\\
\to \frac{{\sqrt x - 1}}{{2\left( {\sqrt x + 1} \right)}} < 0\\
\to \left\{ \begin{array}{l}
\sqrt x + 1 > 0\left( {ld} \right)\forall x \in R\\
\sqrt x - 1 < 0
\end{array} \right.\\
\to \sqrt x < 1\\
\to 0 \le x < 1
\end{array}\)
