Đáp án:
x=0
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne 9\\
P = \dfrac{{3\sqrt x }}{{\sqrt x - 3}}\\
P = {P^4}\\
\Leftrightarrow P - {P^4} = 0\\
\to P\left( {1 - {P^3}} \right) = 0\\
\to \left[ \begin{array}{l}
P = 0\\
{P^3} = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{{3\sqrt x }}{{\sqrt x - 3}} = 0\\
\dfrac{{27.\sqrt {{x^3}} }}{{\sqrt {{x^3}} - 9x + 27\sqrt x - 27}} = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
27.\sqrt {{x^3}} = \sqrt {{x^3}} - 9x + 27\sqrt x - 27
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
26\sqrt {{x^3}} + 9x - 27\sqrt x + 27 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
\sqrt x = - \dfrac{3}{2}\left( l \right)
\end{array} \right.\\
KL:x = 0
\end{array}\)
