Đáp án:
$\begin{array}{l}
b)Dkxd:x \ge \dfrac{1}{2}\\
\sqrt {2x - 1} = 3\\
\Leftrightarrow 2x - 1 = 9\\
\Leftrightarrow 2x = 10\\
\Leftrightarrow x = 5\left( {tmdk} \right)\\
Vậy\,x = 5\\
B2)\\
P = \left( {\dfrac{1}{{\sqrt x + 1}} + \dfrac{{x + \sqrt x + 2}}{{x - 1}}} \right):\dfrac{1}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x - 1 + x + \sqrt x + 2}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\left( {\sqrt x - 1} \right)\\
= \dfrac{{x + 2\sqrt x + 1}}{{\sqrt x + 1}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x + 1}}\\
= \sqrt x + 1\\
b)x = 4\left( {tmdk} \right)\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow P = \sqrt x + 1 = 2 + 1 = 3\\
c)P = 2\\
\Leftrightarrow \sqrt x + 1 = 2\\
\Leftrightarrow \sqrt x = 1\\
\Leftrightarrow x = 1\left( {ktm} \right)
\end{array}$
Vậy ko có x thỏa mãn để $P = 2$
