Đáp án:
\(\dfrac{{2\sqrt x + 3}}{{5\sqrt x + 7}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:\left\{ \begin{array}{l}
\sqrt x \ne \dfrac{2}{3}\\
x \ge 0
\end{array} \right. \to \left\{ \begin{array}{l}
x \ne \dfrac{4}{9}\\
x \ge 0
\end{array} \right.\\
\dfrac{{\left( {3 - \sqrt x } \right)\left( {3\sqrt x - 2} \right) + \left( {3\sqrt x + 4} \right)\left( {5\sqrt x + 7} \right) - 42\sqrt x - 34}}{{\left( {5\sqrt x + 7} \right)\left( {3\sqrt x - 2} \right)}}\\
= \dfrac{{9\sqrt x - 3x - 6 + 2\sqrt x + 15x + 21\sqrt x + 20\sqrt x + 28 - 42\sqrt x - 34}}{{\left( {5\sqrt x + 7} \right)\left( {3\sqrt x - 2} \right)}}\\
= \dfrac{{12x + 10\sqrt x - 12}}{{\left( {5\sqrt x + 7} \right)\left( {3\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {2\sqrt x + 3} \right)\left( {3\sqrt x - 2} \right)}}{{\left( {5\sqrt x + 7} \right)\left( {3\sqrt x - 2} \right)}}\\
= \dfrac{{2\sqrt x + 3}}{{5\sqrt x + 7}}
\end{array}\)
