Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x \ne 0
\end{array} \right. \Leftrightarrow x > 0\\
P = 1 - \sqrt x - \dfrac{{\sqrt x + x}}{{\sqrt x }}\\
= 1 - \sqrt x - \dfrac{{\sqrt x \left( {1 + \sqrt x } \right)}}{{\sqrt x }}\\
= 1 - \sqrt x - \left( {1 + \sqrt x } \right)\\
= 1 - \sqrt x - 1 - \sqrt x \\
= - 2\sqrt x \\
b)P + x - 8 = 0\\
\Leftrightarrow - 2\sqrt x + x - 8 = 0\\
\Leftrightarrow x - 2\sqrt x - 8 = 0\\
\Leftrightarrow x - 4\sqrt x + 2\sqrt x - 8 = 0\\
\Leftrightarrow \left( {\sqrt x - 4} \right)\left( {\sqrt x + 2} \right) = 0\\
\Leftrightarrow \sqrt x = 4\\
\Leftrightarrow x = 16\left( {tmdk} \right)\\
Vậy\,x = 16
\end{array}$
