Đáp án:
2) a. \(\dfrac{x}{{x - 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)a.DK:x \ge 0;x \ne 9\\
P = \left[ {\dfrac{{{{\left( {\sqrt x + 3} \right)}^2} - {{\left( {\sqrt x - 3} \right)}^2}}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}} \right]:\dfrac{6}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x + 6\sqrt x + 9 - x + 6\sqrt x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}.\dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}{6}\\
= \dfrac{{12\sqrt x }}{6} = 2\sqrt x \\
b.Thay:x = 9\\
\to P = 2.\sqrt 9 = 2.3 = 6\\
c.Do:\sqrt x \ge 0\\
\to 2\sqrt x \ge 0\\
\to Min = 0\\
\Leftrightarrow x = 0\\
2)a.DK:x \ne \left\{ { - \sqrt 2 ;0;\sqrt 2 } \right\}\\
P = \left( {\dfrac{{4\sqrt x + 4\sqrt 2 - 3\sqrt x + 3\sqrt 2 }}{{x - 2}}} \right):\dfrac{{x + 7\sqrt 2 }}{x}\\
= \dfrac{{x + 7\sqrt 2 }}{{x - 2}}.\dfrac{x}{{x + 7\sqrt 2 }}\\
= \dfrac{x}{{x - 2}}\\
b.Thay:x = - 1\\
\to P = \dfrac{{ - 1}}{{ - 1 - 2}} = \dfrac{1}{3}\\
c.P = \dfrac{x}{{x - 2}} = \dfrac{{x - 2 + 2}}{{x - 2}} = 1 + \dfrac{2}{{x - 2}}\\
P \in Z \Leftrightarrow \dfrac{2}{{x - 2}} \in Z\\
\Leftrightarrow x - 2 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 2\\
x - 2 = - 2\\
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = 0\left( l \right)\\
x = 3\\
x = 1
\end{array} \right.
\end{array}\)
