Đáp án:
\(\begin{array}{l}
a)\dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
b)0 \le x < 9;x \ne 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
P = \dfrac{{\sqrt x \left( {\sqrt x + 1} \right) + 3\left( {\sqrt x - 1} \right) - 6\sqrt x + 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + \sqrt x + 3\sqrt x - 3 - 6\sqrt x + 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x - 2\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
b)P < \dfrac{1}{2}\\
\to \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} < \dfrac{1}{2}\\
\to \dfrac{{2\sqrt x - 2 - \sqrt x - 1}}{{2\left( {\sqrt x + 1} \right)}} < 0\\
\to \sqrt x - 3 < 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to 0 \le x < 9;x \ne 1
\end{array}\)
