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