Đáp án:
3) \(\left[ \begin{array}{l}
x = 16\\
x = 4\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)P = \dfrac{{x + 1 + \sqrt x }}{{x + 1}}:\left[ {\dfrac{1}{{\sqrt x - 1}} - \dfrac{{2\sqrt x }}{{x\left( {\sqrt x - 1} \right) + \left( {\sqrt x - 1} \right)}}} \right] - \dfrac{{\left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{x - \sqrt x + 1}}\\
= \dfrac{{x + 1 + \sqrt x }}{{x + 1}}:\dfrac{{x + 1 - 2\sqrt x }}{{\left( {x + 1} \right)\left( {\sqrt x - 1} \right)}} - \sqrt x - 1\\
= \dfrac{{x + 1 + \sqrt x }}{{x + 1}}.\dfrac{{\left( {x + 1} \right)\left( {\sqrt x - 1} \right)}}{{{{\left( {\sqrt x - 1} \right)}^2}}} - \sqrt x - 1\\
= \dfrac{{x + \sqrt x + 1}}{{\sqrt x - 1}} - \left( {\sqrt x + 1} \right)\\
= \dfrac{{x + \sqrt x + 1 - x + 1}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}}\\
2)P < 4\\
\to \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}} < 4\\
\to \dfrac{{\sqrt x + 2 - 4\left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}} < 0\\
\to \dfrac{{ - 3\sqrt x + 6}}{{\sqrt x - 1}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
- 3\sqrt x + 6 > 0\\
\sqrt x - 1 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
- 3\sqrt x + 6 < 0\\
\sqrt x - 1 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
1 > \sqrt x \\
\sqrt x > 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
0 \le x < 1\\
x > 4
\end{array} \right.\\
3)P = \dfrac{{\sqrt x + 2}}{{\sqrt x - 1}} = \dfrac{{\sqrt x - 1 + 3}}{{\sqrt x - 1}} = 1 + \dfrac{3}{{\sqrt x - 1}}\\
P \in Z \to \dfrac{3}{{\sqrt x - 1}} \in Z\\
\to \sqrt x - 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 1 = 3\\
\sqrt x - 1 = 1\\
\sqrt x - 1 = - 1\\
\sqrt x - 1 = - 3\left( l \right)
\end{array} \right. \to \left[ \begin{array}{l}
x = 16\\
x = 4\\
x = 0
\end{array} \right.
\end{array}\)
