Đáp án:
$\begin{array}{l}
1)a)Dkxd:x \ge 2\\
\sqrt {9\left( {x - 2} \right)} + \dfrac{3}{2}\sqrt {\dfrac{{4x - 4}}{9}} = 8\\
\Leftrightarrow 3\sqrt {x - 2} + \dfrac{3}{2}.\dfrac{2}{3}\sqrt {x - 2} = 8\\
\Leftrightarrow 3\sqrt {x - 2} + \sqrt {x - 2} = 8\\
\Leftrightarrow 4\sqrt {x - 2} = 8\\
\Leftrightarrow \sqrt {x - 2} = 2\\
\Leftrightarrow x - 2 = 4\\
\Leftrightarrow x = 6\left( {tmdk} \right)\\
Vậy\,x = 6\\
b)Dkxd:x \ge 1\\
\sqrt {6{x^2} + 6x + 1} = x - 1\\
\Leftrightarrow 6{x^2} + 6x + 1 = {x^2} - 2x + 1\\
\Leftrightarrow 5{x^2} + 8x = 0\\
\Leftrightarrow x\left( {5x + 8} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\left( {ktm} \right)\\
x = - \dfrac{8}{5}\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x \in \emptyset \\
2)a)Dkxd:x > 0\\
A = \dfrac{{\sqrt x - 1}}{{\sqrt x }} + \dfrac{1}{{x + \sqrt x }}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - 1 + 1}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{x}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
b)P = A.B\\
B = ??
\end{array}$
