Question

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Answer 1

\[\begin{array}{l}
a)P = \frac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
\left( {x \ge 0;x \ne 9} \right)\\
P \ge - 1 \Leftrightarrow \frac{{\sqrt x + 1}}{{\sqrt x - 3}} \ge - 1\\
\Leftrightarrow \frac{{\sqrt x + 1}}{{\sqrt x - 3}} + 1 \ge 0\\
\Leftrightarrow \frac{{\sqrt x + 1 + \sqrt x - 3}}{{\sqrt x - 3}} \ge 0\\
\Leftrightarrow \frac{{2\sqrt x - 2}}{{\sqrt x - 3}} \ge 0\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
2\sqrt x - 2 \ge 0\\
\sqrt x - 3 > 0
\end{array} \right.}\\
{\left\{ \begin{array}{l}
2\sqrt x - 2 \le 0\\
\sqrt x - 3 < 0
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
\sqrt x \ge 1\\
\sqrt x > 3
\end{array} \right.}\\
{\left\{ \begin{array}{l}
\sqrt x \le 1\\
\sqrt x < 3
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x \ge 1\\
x > 9
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x \le 1\\
x < 9
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x > 9\\
x \le 1
\end{array} \right.\\
vay:1 \le x < 9thiP \ge - 1\\
b)P > 5\\
\Leftrightarrow \frac{{\sqrt x + 1}}{{\sqrt x - 3}} > 5\\
\Leftrightarrow \frac{{\sqrt x + 1}}{{\sqrt x - 3}} - 5 > 0\\
\Leftrightarrow \frac{{\sqrt x + 1 - 5\sqrt x + 15}}{{\sqrt x - 3}} > 0\\
\Leftrightarrow \frac{{16 - 4\sqrt x }}{{\sqrt x - 3}} > 0\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
16 - 4\sqrt x > 0\\
\sqrt x - 3 > 0
\end{array} \right.}\\
{\left\{ \begin{array}{l}
15 - 4\sqrt x < 0\\
\sqrt x - 3 < 0
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
\sqrt x < 4\\
\sqrt x > 3
\end{array} \right.}\\
{\left\{ \begin{array}{l}
\sqrt x > 4\\
\sqrt x < 3
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x < 16\\
x > 9
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x > 16\\
x < 9
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow 9 < x < 16\\
Ma:x \in Z\\
= > x \in \{ 10;11;12;13;14;15\} \\
vay:x \in \{ 10;11;12;13;14;15\} thiP > 5
\end{array}\]

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Answer 2

\[\begin{array}{l}
a)P = \frac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
\left( {x \ge 0;x \ne 9} \right)\\
P \ge - 1 \Leftrightarrow \frac{{\sqrt x + 1}}{{\sqrt x - 3}} \ge - 1\\
\Leftrightarrow \frac{{\sqrt x + 1}}{{\sqrt x - 3}} + 1 \ge 0\\
\Leftrightarrow \frac{{\sqrt x + 1 + \sqrt x - 3}}{{\sqrt x - 3}} \ge 0\\
\Leftrightarrow \frac{{2\sqrt x - 2}}{{\sqrt x - 3}} \ge 0\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
2\sqrt x - 2 \ge 0\\
\sqrt x - 3 > 0
\end{array} \right.}\\
{\left\{ \begin{array}{l}
2\sqrt x - 2 \le 0\\
\sqrt x - 3 < 0
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
\sqrt x \ge 1\\
\sqrt x > 3
\end{array} \right.}\\
{\left\{ \begin{array}{l}
\sqrt x \le 1\\
\sqrt x < 3
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x \ge 1\\
x > 9
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x \le 1\\
x < 9
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x > 9\\
x \le 1
\end{array} \right.\\
vay:1 \le x < 9thiP \ge - 1\\
b)P > 5\\
\Leftrightarrow \frac{{\sqrt x + 1}}{{\sqrt x - 3}} > 5\\
\Leftrightarrow \frac{{\sqrt x + 1}}{{\sqrt x - 3}} - 5 > 0\\
\Leftrightarrow \frac{{\sqrt x + 1 - 5\sqrt x + 15}}{{\sqrt x - 3}} > 0\\
\Leftrightarrow \frac{{16 - 4\sqrt x }}{{\sqrt x - 3}} > 0\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
16 - 4\sqrt x > 0\\
\sqrt x - 3 > 0
\end{array} \right.}\\
{\left\{ \begin{array}{l}
15 - 4\sqrt x < 0\\
\sqrt x - 3 < 0
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
\sqrt x < 4\\
\sqrt x > 3
\end{array} \right.}\\
{\left\{ \begin{array}{l}
\sqrt x > 4\\
\sqrt x < 3
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\left\{ \begin{array}{l}
x < 16\\
x > 9
\end{array} \right.}\\
{\left\{ \begin{array}{l}
x > 16\\
x < 9
\end{array} \right.}
\end{array}} \right.\\
\Leftrightarrow 9 < x < 16\\
Ma:x \in Z\\
= > x \in \{ 10;11;12;13;14;15\} \\
vay:x \in \{ 10;11;12;13;14;15\} thiP > 5
\end{array}\]

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