Đáp án:
\(\begin{array}{l}
a)2\\
b)0 \le x < 9\\
c)\dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
d)x \in \emptyset
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 0;x \ne 9\\
Thay:x = 36\\
\to B = \dfrac{{\sqrt {36} }}{{\sqrt {36} - 3}} = \dfrac{6}{3} = 2\\
b)B < \dfrac{1}{2}\\
\to \dfrac{{\sqrt x }}{{\sqrt x - 3}} < \dfrac{1}{2}\\
\to \dfrac{{2\sqrt x - \sqrt x + 3}}{{2\left( {\sqrt x - 3} \right)}} < 0\\
\to \dfrac{{\sqrt x + 3}}{{2\left( {\sqrt x - 3} \right)}} < 0\\
\to \sqrt x - 3 < 0\left( {do:\sqrt x + 3 > 0\forall x \ge 0} \right)\\
\to 0 \le x < 9\\
c)DK:x > 0;x \ne 1\\
A = \dfrac{{x - 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}} = \dfrac{{\sqrt x + 1}}{{\sqrt x }}\\
d)P = \dfrac{1}{{A.B}} = \dfrac{1}{{\dfrac{{\sqrt x + 1}}{{\sqrt x }}.\dfrac{{\sqrt x }}{{\sqrt x - 3}}}}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 - 4}}{{\sqrt x + 1}}\\
= 1 - \dfrac{4}{{\sqrt x + 1}}\\
P \in Z \to \dfrac{4}{{\sqrt x + 1}} \in Z \to \sqrt x + 1 \in U\left( 4 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 4\\
\sqrt x + 1 = 2\\
\sqrt x + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\left( l \right)\\
x = 1\left( l \right)\\
x = 0\left( l \right)
\end{array} \right.\\
\to x \in \emptyset
\end{array}\)
