Đáp án:
$\begin{array}{l}
a)Dkxd:x > 0;x \ne 1\\
P = \left( {\dfrac{{x - 2}}{{x + 2\sqrt x }} + \dfrac{1}{{\sqrt x + 2}}} \right).\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{x - 2 + \sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {\sqrt x + 2} \right)}}.\dfrac{{\sqrt x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
b)2P = 2\sqrt x + 5\\
\Rightarrow \dfrac{{2\sqrt x + 2}}{{\sqrt x }} = 2\sqrt x + 5\\
\Rightarrow 2\sqrt x + 2 = 2x + 5\sqrt x \\
\Rightarrow 2x + 3\sqrt x - 2 = 0\\
\Rightarrow \left( {2\sqrt x - 1} \right)\left( {\sqrt x + 2} \right) = 0\\
\Rightarrow \sqrt x = \dfrac{1}{2}\left( {do:\sqrt x + 2 > 0} \right)\\
\Rightarrow x = \dfrac{1}{4}\left( {tmdk} \right)\\
C3)\\
1)A\left( { - \dfrac{1}{2};\dfrac{2}{3}} \right) \in y = \left( {2m - 3} \right).x - \dfrac{1}{2}\\
\Rightarrow \dfrac{2}{3} = \left( {2m - 3} \right).\left( {\dfrac{{ - 1}}{2}} \right) - \dfrac{1}{2}\\
\Rightarrow 4 = - 3.\left( {2m - 3} \right) - 3\\
\Rightarrow - 6m + 9 - 3 = 4\\
\Rightarrow 6m = 2\\
\Rightarrow m = \dfrac{1}{3}\\
2){x^2} - 2x - 2m + 1 = 0\\
\Rightarrow \Delta ' > 0\\
\Rightarrow 1 + 2m - 1 > 0\\
\Rightarrow m > 0\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = - 2m + 1
\end{array} \right.\\
x_2^2\left( {x_1^2 - 1} \right) + x_1^2\left( {x_2^2 - 1} \right) = 8\\
\Rightarrow 2{\left( {{x_1}{x_2}} \right)^2} - \left( {x_1^2 + x_2^2} \right) = 8\\
\Rightarrow 2.{\left( { - 2m + 1} \right)^2} - {\left( {{x_1} + {x_2}} \right)^2} + 2{x_1}{x_2} = 8\\
\Rightarrow 2.{\left( {2m - 1} \right)^2} - 4 + 2\left( { - 2m + 1} \right) - 8 = 0\\
\Rightarrow {\left( {2m - 1} \right)^2} - \left( {2m - 1} \right) - 6 = 0\\
\Rightarrow \left( {2m - 1 - 3} \right)\left( {2m - 1 + 2} \right) = 0\\
\Rightarrow \left( {2m - 4} \right)\left( {2m + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 2\left( {tm} \right)\\
m = - \dfrac{1}{2}\left( {ktm} \right)
\end{array} \right.\\
Vậy\,m = 2
\end{array}$
