Đáp án:
c) x=1
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)Thay:x = \dfrac{1}{9}\\
\to A = \dfrac{2}{{\sqrt {\dfrac{1}{9}} - 3}} = - \dfrac{3}{4}\\
b)B = \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 3} \right) - \sqrt x \left( {\sqrt x + 3} \right) + 9\sqrt x - 5}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x - 4\sqrt x + 3 - x - 3\sqrt x + 9\sqrt x - 5}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2\sqrt x - 2}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\
P = B:A = \dfrac{{2\sqrt x - 2}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}:\dfrac{2}{{\sqrt x - 3}}\\
= \dfrac{{2\sqrt x - 2}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}.\dfrac{{\sqrt x - 3}}{2}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 3}}\\
c)P = \dfrac{{\sqrt x - 1}}{{\sqrt x + 3}} = \dfrac{{\sqrt x + 3 - 4}}{{\sqrt x + 3}} = 1 - \dfrac{4}{{\sqrt x + 3}}\\
P \in Z \Leftrightarrow \dfrac{4}{{\sqrt x + 3}} \in Z\\
\Leftrightarrow \sqrt x + 3 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 3 = 4\\
\sqrt x + 3 = 2\left( l \right)\\
\sqrt x + 3 = 1\left( l \right)
\end{array} \right.\\
\to \sqrt x = 1\\
\to x = 1
\end{array}\)
