Đáp án:
a) \(\dfrac{{3\sqrt x + 1}}{{\sqrt x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 1\\
P = \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} + {{\left( {\sqrt x - 1} \right)}^2} - 3\sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{x + 2\sqrt x + 1 + x - 2\sqrt x + 1 - 3\sqrt x - 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{2x - 3\sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {3\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{{3\sqrt x + 1}}{{\sqrt x + 1}}\\
b)Thay:x = 9\\
\to P = \dfrac{{3\sqrt 9 + 1}}{{\sqrt 9 + 1}} = \dfrac{{10}}{4} = \dfrac{5}{2}\\
c)P = \dfrac{1}{2}\\
\to \dfrac{{3\sqrt x + 1}}{{\sqrt x + 1}} = \dfrac{1}{2}\\
\to 6\sqrt x + 2 = \sqrt x + 1\\
\to 5\sqrt x = - 1\left( {vô lý} \right)\\
\to x \in \emptyset \\
d)P = \dfrac{{3\sqrt x + 1}}{{\sqrt x + 1}} = \dfrac{{3\left( {\sqrt x + 1} \right) - 2}}{{\sqrt x + 1}}\\
= 3 - \dfrac{2}{{\sqrt x + 1}}\\
P \in Z \Leftrightarrow \dfrac{2}{{\sqrt x + 1}} \in Z\\
\Leftrightarrow \sqrt x + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 2\\
\sqrt x + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = 0\left( {TM} \right)
\end{array} \right.\\
e)P < 1\\
\to \dfrac{{3\sqrt x + 1}}{{\sqrt x + 1}} < 1\\
\to \dfrac{{3\sqrt x + 1 - \sqrt x - 1}}{{\sqrt x + 1}} < 0\\
\to 2\sqrt x < 0\left( {do:\sqrt x + 1 > 0\forall x \ge 0} \right)\\
\to \sqrt x < 0\left( {vô lý} \right)\\
\to x \in \emptyset
\end{array}\)
