Đáp án:
2) c. x=9
Giải thích các bước giải:
\(\begin{array}{l}
1)\dfrac{{3 - \sqrt 2 }}{{3 + \sqrt 2 }} + \dfrac{{3 + \sqrt 2 }}{{3 - \sqrt 2 }}\\
= \dfrac{{{{\left( {3 - \sqrt 2 } \right)}^2} + {{\left( {3 + \sqrt 2 } \right)}^2}}}{{\left( {3 + \sqrt 2 } \right)\left( {3 - \sqrt 2 } \right)}}\\
= \dfrac{{9 - 6\sqrt 2 + 2 + 9 + 6\sqrt 2 + 2}}{{9 - 2}}\\
= \dfrac{{22}}{7}\\
2)DK:x \ge 0;x \ne 1\\
a.P = \left[ {1 - \dfrac{{4\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} + \dfrac{1}{{\sqrt x - 1}}} \right]:\dfrac{{x - 2\sqrt x }}{{x - 1}}\\
= \left[ {\dfrac{{x - 1 - 4\sqrt x + \sqrt x + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}} \right].\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x - 3\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x - 2}}\\
b.Thay:x = 16\\
\to P = \dfrac{{\sqrt {16} - 3}}{{\sqrt {16} - 2}} = \dfrac{{4 - 3}}{{4 - 2}} = \dfrac{1}{2}\\
c.P = \dfrac{{\sqrt x - 2 - 1}}{{\sqrt x - 2}} = 1 - \dfrac{1}{{\sqrt x - 2}}\\
P \in Z \Leftrightarrow \dfrac{1}{{\sqrt x - 2}} \in Z\\
\Leftrightarrow \sqrt x - 2 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
\sqrt x - 2 = 1\\
\sqrt x - 2 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 3\\
\sqrt x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
x = 1\left( l \right)
\end{array} \right.
\end{array}\)
