Đáp án:
b. \(x > 1;x \ne 9\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x > 0;x \ne \left\{ {1;9} \right\}\\
P = \dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}{{1 - x}}.\left[ {\dfrac{{\sqrt x - 3 + \sqrt x + 3}}{{\sqrt x \left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}} \right]\\
= \dfrac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}{{1 - x}}.\dfrac{{2\sqrt x }}{{\sqrt x \left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{2}{{1 - x}}\\
b.P < 1\\
\to \dfrac{2}{{1 - x}} < 1\\
\to \dfrac{{2 - 1 + x}}{{1 - x}} < 0\\
\to \dfrac{{1 + x}}{{1 - x}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 1 > 0\\
1 - x < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 1 < 0\\
1 - x > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - 1\\
x > 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - 1\\
x < 1
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > 1\\
x < - 1
\end{array} \right.
\end{array}\)
Kết hợp ĐK ⇒ \(x > 1;x \ne 9\)
