Đáp án:
\(\begin{array}{l}
a)\dfrac{{3\sqrt x }}{{\sqrt x + 3}}\\
b)\sqrt 3 \\
c)x \in \emptyset
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)P = \dfrac{{\sqrt x }}{{\sqrt x + 3}} + \dfrac{{3x + 9}}{{9 - x}} + \dfrac{{2\sqrt x }}{{\sqrt x - 3}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 3} \right) - 3x - 9 + 2\sqrt x \left( {\sqrt x + 3} \right)}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 3\sqrt x - 3x - 9 + 2x + 6\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{3\sqrt x - 9}}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 3}}\\
b)Thay:x = 4 + 2\sqrt 3 \\
= 3 + 2\sqrt 3 .1 + 1\\
= {\left( {\sqrt 3 + 1} \right)^2}\\
\to Q = \dfrac{3}{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - 1}} = \dfrac{3}{{\sqrt 3 + 1 - 1}}\\
= \sqrt 3 \\
c)Q:P = \dfrac{3}{{\sqrt x - 1}}:\dfrac{{3\sqrt x }}{{\sqrt x + 3}}\\
= \dfrac{3}{{\sqrt x - 1}}.\dfrac{{\sqrt x + 3}}{{3\sqrt x }} = \dfrac{{\sqrt x + 3}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x - 1 + 4}}{{\sqrt x \left( {\sqrt x - 1} \right)}} = \dfrac{1}{{\sqrt x }} + \dfrac{4}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
Q:P > 0\\
\Leftrightarrow \dfrac{{\sqrt x + 3}}{{\sqrt x \left( {\sqrt x - 1} \right)}} > 0\\
\to \sqrt x - 1 > 0\left( {do:x > 0} \right)\\
\to x > 1;x \ne 9\\
Q:P \in Z \to \left\{ \begin{array}{l}
\dfrac{1}{{\sqrt x }} \in Z\\
\dfrac{4}{{\sqrt x \left( {\sqrt x - 1} \right)}} \in Z
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\left( {KTM} \right)\\
\dfrac{4}{{\sqrt x \left( {\sqrt x - 1} \right)}} \in Z
\end{array} \right.\\
\to x \in \emptyset
\end{array}\)
