Đáp án:
\(\begin{array}{l}
1)x \in \emptyset \\
2)\left[ \begin{array}{l}
x = 0\\
x = 25
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ge 0;x \ne \left\{ {4;9} \right\}\\
B = \dfrac{{x - 5 - 3\left( {\sqrt x - 2} \right) - \sqrt x + 3}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 2 - \sqrt x - 3\sqrt x + 6}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 4\sqrt x + 4}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x - 3}}\\
P = A.B = \dfrac{{\sqrt x - 3}}{{\sqrt x + 1}}.\dfrac{{\sqrt x - 2}}{{\sqrt x - 3}}\\
= \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}}\\
\left| P \right| + P = 0\\
\to \left[ \begin{array}{l}
P + P = 0\\
- P + P = 0\left( {ld} \right)
\end{array} \right.\\
\to P = 0\\
\to \dfrac{{\sqrt x - 2}}{{\sqrt x + 1}} = 0\\
\to \sqrt x - 2 = 0\\
\to x = 4\left( {KTM} \right)\\
\to x \in \emptyset \\
2)\left( {x - 1} \right).A = 2x - 9\sqrt x + 3\\
\to \left( {x - 1} \right).\dfrac{{\sqrt x - 3}}{{\sqrt x + 1}} = 2x - 9\sqrt x + 3\\
\to \left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right) = 2x - 9\sqrt x + 3\\
\to x - 4\sqrt x + 3 = 2x - 9\sqrt x + 3\\
\to x - 5\sqrt x = 0\\
\to \sqrt x \left( {\sqrt x - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 25
\end{array} \right.
\end{array}\)
