A=( √x ÷ 3+ √x+x+9 ÷9-x) ÷ (3 √x +1 ÷ x-3 √x – 1 ÷ √x ) a) tìm x để a xác định b)rút gọn x

Giải thích các bước giải:

\(\begin{array}{l}
A = \left( {\dfrac{{\sqrt x }}{{3 + \sqrt x }} + \dfrac{{x + 9}}{{9 – x}}} \right):\left( {\dfrac{{3\sqrt x  + 1}}{{x – 3\sqrt x }} – \dfrac{1}{{\sqrt x }}} \right)\\
a,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
3 + \sqrt x  \ne 0\\
9 – x \ne 0\\
x – 3\sqrt x  \ne 0\\
\sqrt x  \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 0\\
x \ne 9
\end{array} \right.\\
b,\\
A = \left( {\dfrac{{\sqrt x }}{{3 + \sqrt x }} + \dfrac{{x + 9}}{{9 – x}}} \right):\left( {\dfrac{{3\sqrt x  + 1}}{{x – 3\sqrt x }} – \dfrac{1}{{\sqrt x }}} \right)\\
 = \left( {\dfrac{{\sqrt x }}{{3 + \sqrt x }} + \dfrac{{x + 9}}{{\left( {3 – \sqrt x } \right)\left( {3 + \sqrt x } \right)}}} \right):\left( {\dfrac{{3\sqrt x  + 1}}{{\sqrt x \left( {\sqrt x  – 3} \right)}} – \dfrac{1}{{\sqrt x }}} \right)\\
 = \dfrac{{\sqrt x \left( {3 – \sqrt x } \right) + x + 9}}{{\left( {3 – \sqrt x } \right)\left( {3 + \sqrt x } \right)}}:\dfrac{{\left( {3\sqrt x  + 1} \right) – \left( {\sqrt x  – 3} \right)}}{{\sqrt x \left( {\sqrt x  – 3} \right)}}\\
 = \dfrac{{3\sqrt x  – x + x + 9}}{{\left( {3 – \sqrt x } \right)\left( {3 + \sqrt x } \right)}}:\dfrac{{2\sqrt x  + 4}}{{\sqrt x \left( {\sqrt x  – 3} \right)}}\\
 = \dfrac{{3\sqrt x  + 9}}{{\left( {3 – \sqrt x } \right)\left( {3 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  – 3} \right)}}{{2\sqrt x  + 4}}\\
 = \dfrac{{3.\left( {\sqrt x  + 3} \right)}}{{\left( {3 – \sqrt x } \right)\left( {3 + \sqrt x } \right)}}.\dfrac{{\sqrt x \left( {\sqrt x  – 3} \right)}}{{2\sqrt x  + 4}}\\
 = \dfrac{{ – 3\sqrt x }}{{2\sqrt x  + 4}}
\end{array}\)